How to create a polynomial to model the sum of the faces on a cube?
$begingroup$
I am given these three problems:
I think I understand the first question, it is basically asking me to find the formula for the sum of the first odd $n$ cubes, correct? Basically, I can use the finite differences method. So I have:
$1^3+3^3 +5^3+...+(2n-1)^3$
So I have the sequence of:
$1,28,153,496,1225,2556,...$
The first difference gives me: $27, 125, 343, 729, 1331,...$
Second difference is: $98, 218, 386, 602,... $
Third difference is: $120, 168, 216,...$
Fourth difference is: $48, 48,...$
The fourth difference is constant so the polynomial is of the form:
$Ax^4+Bx^3+Cx^2+Dx+E=y$
So from here I can create a system of equations and solve, and I know how to do this.
I know (from google) that the answer is $n^2(2n^2-1)$ but with my method of creating a system of equations and solving for$A,B,C,D,E$, I will get this formula right?
For the second and third questions I'm pretty confused.
For level $n=1$ I have that the number of exposed faces is 5.
$n=2, f= 20$
$n=3, f = 36$
$n=4, f = 52$
Maybe my notation is a little confusing because based off of
$1^3+3^3+...+(2n-1)^3$, $n$ can only be the odd numbers, so should I use a different variable, maybe like $y$? My notation is making me even more confused.
I found a pattern for finding the number of exposed faces (if my interpretation of the question is correct). Namely, the formula for the number of $n$ exposed faces is $4(2n-1)$
Now I am basically asked to find a formula which models the sum of these, so like $5+20+36+52+68+...+4(2n-1) = S_n$ where $S_n$ is the sum.
So I do the same thing as my first question and have that my sequence is:
$5, 20, 36, 52, 68, 84,...$
First difference is: $15, 16, 16, 16, 16, ...$
Second: $1, 0, 0, 0,...$
Third: $-1, 0, 0,...$
Fourth: $1,0,0,0,...$
The first term keeps oscillating between $1$ and $-1$ and while this is cool, it is also really annoying because I don't know why this is happening or if I did something wrong.
EDIT: I realized what I did wrong:
My sequence should be: $5,25,61, 113, 181, 265$
First difference: $20,36,52,68,84,...$
Second: $16,16,16,16,...$
So again using finite differences I should have $$8n^2-4n+1$$
Is this correct?
Additionally, for the third question, what would I have to do to erase the faces on the bottom of the pyramid? This really confuses me because i thought that implied in the assumption that we are dealing with the exposed faces, we would automatically discount the bottom row for the $n$th level, so I don't even see a difference in question 2 and 3.
Any help is appreciated, thank you!
EDIT: I'm thinking of the distinction between question 2 and 3
So The difference would be adding n x n = $n^2$ faces to whatever sum is of the exposed faces. So should the answers for questions 2 and 3 be:
2) $9n^2-4n+1$
3) $8n^2-4n+1$
polynomials
asked Nov 27 '18 at 19:03
user8290579user8290579
37318
$endgroup$
add a comment |
$begingroup$
I am given these three problems:
I think I understand the first question, it is basically asking me to find the formula for the sum of the first odd $n$ cubes, correct? Basically, I can use the finite differences method. So I have:
$1^3+3^3 +5^3+...+(2n-1)^3$
So I have the sequence of:
$1,28,153,496,1225,2556,...$
The first difference gives me: $27, 125, 343, 729, 1331,...$
Second difference is: $98, 218, 386, 602,... $
Third difference is: $120, 168, 216,...$
Fourth difference is: $48, 48,...$
The fourth difference is constant so the polynomial is of the form:
$Ax^4+Bx^3+Cx^2+Dx+E=y$
So from here I can create a system of equations and solve, and I know how to do this.
I know (from google) that the answer is $n^2(2n^2-1)$ but with my method of creating a system of equations and solving for$A,B,C,D,E$, I will get this formula right?
For the second and third questions I'm pretty confused.
For level $n=1$ I have that the number of exposed faces is 5.
$n=2, f= 20$
$n=3, f = 36$
$n=4, f = 52$
Maybe my notation is a little confusing because based off of
$1^3+3^3+...+(2n-1)^3$, $n$ can only be the odd numbers, so should I use a different variable, maybe like $y$? My notation is making me even more confused.
I found a pattern for finding the number of exposed faces (if my interpretation of the question is correct). Namely, the formula for the number of $n$ exposed faces is $4(2n-1)$
Now I am basically asked to find a formula which models the sum of these, so like $5+20+36+52+68+...+4(2n-1) = S_n$ where $S_n$ is the sum.
So I do the same thing as my first question and have that my sequence is:
$5, 20, 36, 52, 68, 84,...$
First difference is: $15, 16, 16, 16, 16, ...$
Second: $1, 0, 0, 0,...$
Third: $-1, 0, 0,...$
Fourth: $1,0,0,0,...$
The first term keeps oscillating between $1$ and $-1$ and while this is cool, it is also really annoying because I don't know why this is happening or if I did something wrong.
EDIT: I realized what I did wrong:
My sequence should be: $5,25,61, 113, 181, 265$
First difference: $20,36,52,68,84,...$
Second: $16,16,16,16,...$
So again using finite differences I should have $$8n^2-4n+1$$
Is this correct?
Additionally, for the third question, what would I have to do to erase the faces on the bottom of the pyramid? This really confuses me because i thought that implied in the assumption that we are dealing with the exposed faces, we would automatically discount the bottom row for the $n$th level, so I don't even see a difference in question 2 and 3.
Any help is appreciated, thank you!
EDIT: I'm thinking of the distinction between question 2 and 3
So The difference would be adding n x n = $n^2$ faces to whatever sum is of the exposed faces. So should the answers for questions 2 and 3 be:
2) $9n^2-4n+1$
3) $8n^2-4n+1$
polynomials
asked Nov 27 '18 at 19:03
user8290579user8290579
37318
$endgroup$
$begingroup$
Your answer for 3) is correct, for 2) you made the error of assuming the bottom layer is $ntimes n$, but it is $(2n-1)times(2n-1)$. Of course, the arguments you gave may convince you that the forumulas are correct, but the idea that the formula is some kind of polynom and you can find it via the difference scheme is not bad but also not convincing to somebody who has never seen such a problem. For 1) my comment from the answer below still stands: Summing 3rd powers is not the right idea.
$endgroup$
– Ingix
Nov 28 '18 at 9:52
add a comment |
$begingroup$
I am given these three problems:
I think I understand the first question, it is basically asking me to find the formula for the sum of the first odd $n$ cubes, correct? Basically, I can use the finite differences method. So I have:
$1^3+3^3 +5^3+...+(2n-1)^3$
So I have the sequence of:
$1,28,153,496,1225,2556,...$
The first difference gives me: $27, 125, 343, 729, 1331,...$
Second difference is: $98, 218, 386, 602,... $
Third difference is: $120, 168, 216,...$
Fourth difference is: $48, 48,...$
The fourth difference is constant so the polynomial is of the form:
$Ax^4+Bx^3+Cx^2+Dx+E=y$
So from here I can create a system of equations and solve, and I know how to do this.
I know (from google) that the answer is $n^2(2n^2-1)$ but with my method of creating a system of equations and solving for$A,B,C,D,E$, I will get this formula right?
For the second and third questions I'm pretty confused.
For level $n=1$ I have that the number of exposed faces is 5.
$n=2, f= 20$
$n=3, f = 36$
$n=4, f = 52$
Maybe my notation is a little confusing because based off of
$1^3+3^3+...+(2n-1)^3$, $n$ can only be the odd numbers, so should I use a different variable, maybe like $y$? My notation is making me even more confused.
I found a pattern for finding the number of exposed faces (if my interpretation of the question is correct). Namely, the formula for the number of $n$ exposed faces is $4(2n-1)$
Now I am basically asked to find a formula which models the sum of these, so like $5+20+36+52+68+...+4(2n-1) = S_n$ where $S_n$ is the sum.
So I do the same thing as my first question and have that my sequence is:
$5, 20, 36, 52, 68, 84,...$
First difference is: $15, 16, 16, 16, 16, ...$
Second: $1, 0, 0, 0,...$
Third: $-1, 0, 0,...$
Fourth: $1,0,0,0,...$
The first term keeps oscillating between $1$ and $-1$ and while this is cool, it is also really annoying because I don't know why this is happening or if I did something wrong.
EDIT: I realized what I did wrong:
My sequence should be: $5,25,61, 113, 181, 265$
First difference: $20,36,52,68,84,...$
Second: $16,16,16,16,...$
So again using finite differences I should have $$8n^2-4n+1$$
Is this correct?
Additionally, for the third question, what would I have to do to erase the faces on the bottom of the pyramid? This really confuses me because i thought that implied in the assumption that we are dealing with the exposed faces, we would automatically discount the bottom row for the $n$th level, so I don't even see a difference in question 2 and 3.
Any help is appreciated, thank you!
EDIT: I'm thinking of the distinction between question 2 and 3
So The difference would be adding n x n = $n^2$ faces to whatever sum is of the exposed faces. So should the answers for questions 2 and 3 be:
2) $9n^2-4n+1$
3) $8n^2-4n+1$
polynomials
asked Nov 27 '18 at 19:03
user8290579user8290579
37318
$endgroup$
I am given these three problems:
I think I understand the first question, it is basically asking me to find the formula for the sum of the first odd $n$ cubes, correct? Basically, I can use the finite differences method. So I have:
$1^3+3^3 +5^3+...+(2n-1)^3$
So I have the sequence of:
$1,28,153,496,1225,2556,...$
The first difference gives me: $27, 125, 343, 729, 1331,...$
Second difference is: $98, 218, 386, 602,... $
Third difference is: $120, 168, 216,...$
Fourth difference is: $48, 48,...$
The fourth difference is constant so the polynomial is of the form:
$Ax^4+Bx^3+Cx^2+Dx+E=y$
So from here I can create a system of equations and solve, and I know how to do this.
I know (from google) that the answer is $n^2(2n^2-1)$ but with my method of creating a system of equations and solving for$A,B,C,D,E$, I will get this formula right?
For the second and third questions I'm pretty confused.
For level $n=1$ I have that the number of exposed faces is 5.
$n=2, f= 20$
$n=3, f = 36$
$n=4, f = 52$
Maybe my notation is a little confusing because based off of
$1^3+3^3+...+(2n-1)^3$, $n$ can only be the odd numbers, so should I use a different variable, maybe like $y$? My notation is making me even more confused.
I found a pattern for finding the number of exposed faces (if my interpretation of the question is correct). Namely, the formula for the number of $n$ exposed faces is $4(2n-1)$
Now I am basically asked to find a formula which models the sum of these, so like $5+20+36+52+68+...+4(2n-1) = S_n$ where $S_n$ is the sum.
So I do the same thing as my first question and have that my sequence is:
$5, 20, 36, 52, 68, 84,...$
First difference is: $15, 16, 16, 16, 16, ...$
Second: $1, 0, 0, 0,...$
Third: $-1, 0, 0,...$
Fourth: $1,0,0,0,...$
The first term keeps oscillating between $1$ and $-1$ and while this is cool, it is also really annoying because I don't know why this is happening or if I did something wrong.
EDIT: I realized what I did wrong:
My sequence should be: $5,25,61, 113, 181, 265$
First difference: $20,36,52,68,84,...$
Second: $16,16,16,16,...$
So again using finite differences I should have $$8n^2-4n+1$$
Is this correct?
Additionally, for the third question, what would I have to do to erase the faces on the bottom of the pyramid? This really confuses me because i thought that implied in the assumption that we are dealing with the exposed faces, we would automatically discount the bottom row for the $n$th level, so I don't even see a difference in question 2 and 3.
Any help is appreciated, thank you!
EDIT: I'm thinking of the distinction between question 2 and 3
So The difference would be adding n x n = $n^2$ faces to whatever sum is of the exposed faces. So should the answers for questions 2 and 3 be:
2) $9n^2-4n+1$
3) $8n^2-4n+1$
polynomials
polynomials
asked Nov 27 '18 at 19:03
user8290579user8290579
37318
asked Nov 27 '18 at 19:03
user8290579user8290579
37318
edited Nov 27 '18 at 19:21
user8290579
asked Nov 27 '18 at 19:03
user8290579user8290579
37318
asked Nov 27 '18 at 19:03
user8290579user8290579
37318
asked Nov 27 '18 at 19:03
user8290579user8290579
37318
37318
$begingroup$
Your answer for 3) is correct, for 2) you made the error of assuming the bottom layer is $ntimes n$, but it is $(2n-1)times(2n-1)$. Of course, the arguments you gave may convince you that the forumulas are correct, but the idea that the formula is some kind of polynom and you can find it via the difference scheme is not bad but also not convincing to somebody who has never seen such a problem. For 1) my comment from the answer below still stands: Summing 3rd powers is not the right idea.
$endgroup$
– Ingix
Nov 28 '18 at 9:52
add a comment |
$begingroup$
Your answer for 3) is correct, for 2) you made the error of assuming the bottom layer is $ntimes n$, but it is $(2n-1)times(2n-1)$. Of course, the arguments you gave may convince you that the forumulas are correct, but the idea that the formula is some kind of polynom and you can find it via the difference scheme is not bad but also not convincing to somebody who has never seen such a problem. For 1) my comment from the answer below still stands: Summing 3rd powers is not the right idea.
$endgroup$
– Ingix
Nov 28 '18 at 9:52
$begingroup$
Your answer for 3) is correct, for 2) you made the error of assuming the bottom layer is $ntimes n$, but it is $(2n-1)times(2n-1)$. Of course, the arguments you gave may convince you that the forumulas are correct, but the idea that the formula is some kind of polynom and you can find it via the difference scheme is not bad but also not convincing to somebody who has never seen such a problem. For 1) my comment from the answer below still stands: Summing 3rd powers is not the right idea.
$endgroup$
– Ingix
Nov 28 '18 at 9:52
$begingroup$
Your answer for 3) is correct, for 2) you made the error of assuming the bottom layer is $ntimes n$, but it is $(2n-1)times(2n-1)$. Of course, the arguments you gave may convince you that the forumulas are correct, but the idea that the formula is some kind of polynom and you can find it via the difference scheme is not bad but also not convincing to somebody who has never seen such a problem. For 1) my comment from the answer below still stands: Summing 3rd powers is not the right idea.
$endgroup$
– Ingix
Nov 28 '18 at 9:52
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
It's hard to know what you are counting and what you are not counting.
You have three different questions; one can see when you stop discussing the first one and start on the other two, but you do not clearly state when you are working on part 2 and when on part 3.
Your use of the formula $4(2n-1)$ is confusing, because in this formula you decided to ignore the meaning of $n$ as given in the problem statement,
where $n$ is the number of levels
(which can have values $1, 2, 3, 4, ldots$).
Coincidentally, $4(2n-1)$ happens to be the number of faces on the $n$th level that are exposed on the vertical sides, not including the faces exposed on the top or bottom of any cube.
Instead of staying consistent with the notation in the problem statement,
you decided to use the variable $n$ to represent something different,
a number that can only take odd values.
You get lucky later because by the time you are trying to derive a formula
for the "real" value of $n$ as defined in the problem statement,
you have only explicit integers like $113$ which have "forgotten" your other definition of $n.$
An approach that would be less confusing for readers who aren't inside your head would be to introduce a new variable for the odd numbers you wanted to represent, for example $m.$
Then you are free to say $m$ has only odd values without contradicting anything in the problem statement, and you can write $4(2m-1)$ for the number of new faces added by the last level, where $m$ is the number of cubes along one edge of that level, which is always an odd number.
Note that if you define $m$ this way, then $m = 2n - 1$ where $n$ is the number of levels added so far.
At the end you got the correct sequence for part 3.
One may guess that you did this by taking partial sums of the series
$5 + 20 + 36 + 36 + 52 + 68 + cdots.$
It would be better if you said you were doing that.
After that point, although you did not provide all the details of the finite difference method, at least you said you were using finite differences so we know what you were trying to do there.
The difference between questions 2 and 3 has to do with the the implicit assumption that we are always dealing with a pyramid that is finite in size
(though it can be arbitrarily large, not limited to $11$ levels like the pyramid in the figure)
and also with the implicit assumption that
"the entire pyramid up to level $n$"
is a pyramid of exactly $n$ levels that you can look at from all angles, so that your ability to look at the bottom of the $n$th level of the pyramid is not blocked by being stuck to some even larger level underneath it.
To be fair, I think the problem would be much better worded if it explicitly said "a pyramid of $n$ levels."
If it were really a problem about the top $n$ levels of the exact pyramid shown in the figure, which has $11$ levels, then the formula for the number of exposed faces would be identical for questions 2 and 3 when $n < 11$ but would suddenly become much larger for question 2 when $n = 11.$
Also, the answers for $n > 11$ would all be numerically the same as for
$n = 11$; that is, since there are only $11$ levels, we don't find any more sides by attempting to count more than $11$ levels.
This would make a very silly problem, which is why I did not even consider this interpretation at first but only considered the interpretation where $n$ is the total number of levels in an arbitrary pyramid.
As an alternative approach, which I think is easier (although not useful as an exercise in using the method of finite differences),
you could look at the pyramid's orthogonal views from each of the six directions: above, below, left, right, front, back.
In each direction you see a set of exposed faces of cubes--a different set of faces in each direction, and all exposed faces are seen in one of the directions,
so all you have to do is find the number of faces seen from each direction
and add up those six numbers
(or in question 3, five numbers, since we do not include the number of faces on the bottom).
To make it even easier, the top and bottom views are identical large squares
of edge length $2n - 1$,
and the other four views are all arrangements of the form
$1 + 3 + 5 + cdots + (2n - 1)$ for $n$ layers,
which you should probably recognize and know how to express as a polynomial.
Using this approach one can quickly confirm that the formula you found,
$8n^2 - 4n + 1,$ is the correct answer for question 3.
We have $(2n - 1)^2 = 4n^2 - 4 + 1$ faces visible from above,
and $n^2$ faces visible from each of the other four views
(excluding the bottom view).
answered Nov 28 '18 at 4:42
David KDavid K
53.9k342116
$endgroup$
$begingroup$
The way that I got 4(2n-1) ad my formula is because my values for n are all odd numbers. For example, $n=3: 4(5)=20$ which is the correct value for the number of exposed faces for the 3 x 3. I explained how I got the answer for question 3, I created a sequence of numbers which represent the sum of all the exposed faces, and then did the finite differences method to find a polynomial which matches it.
$endgroup$
– user8290579
Nov 28 '18 at 14:30
$begingroup$
thank you very much for explaining the $4(2m-1), m = 2n-1$ I was really confused about that but this makes it a lot clearer and it would make it a lot clearer to other people. unfortunately the only method i can use is the finite differences method. and for the distinction between problems 2 and 3, you are basically saying that for problem 2, including the bottom faces means excluding the bottom faces for (n-1) levels and including it for the nth level, right? and for question 3, we are excluding the bottom faces for all n levels, right? Why am i getting correct answers w the formulas i have
$endgroup$
– user8290579
Nov 28 '18 at 16:42
$begingroup$
You are right, in question 2 the bottom faces are counted only on the $n$th level. It occurred to me (after writing the initial answer) that this might be intended as an exercise in finite differences, so solving it the "easy" way would get you no credit.
$endgroup$
– David K
Nov 28 '18 at 18:07
$begingroup$
Regarding the confusion over $4(n-1),$ notice that if you changed $n$ to $m$ just in the two paragraphs that mention $4(n-1),$ you don't have to change anything else to make sense of it. I think you just had (in your mind) a temporary definition of $n$ for those two paragraphs, and you used it to get the sequence $5,25,61,113,181,265,$ which is a correct sequence for part 3. Notice there is no $n$ in that sequence; it doesn't use the "temporary" definition of $n$ and you never go back to that definition. The next time $n$ appears you're using it exactly as intended in the problem statement.
$endgroup$
– David K
Nov 28 '18 at 18:13
1
$begingroup$
Got it, so if I add $4n^2-4n+1$ to the sequence $8n^2-4n+1$ then that would be the correct answer?
$endgroup$
– user8290579
Nov 28 '18 at 23:22
|
show 2 more comments
$begingroup$
It's a bit much to comment on everything, but
for question 1): You are summing cubes (3rd powers), but that is not correct. Note that the number of cubes in the second layer from the top is $9=3^2$, nor $27=3^3$ as you wrote. The same goes of course for the other layers.
For question 2), 3): Based on your answer for $n=1$, you seem to be considering question 3 (no bottom faces). But I get 21 as the value for $n=2$, not 20 or 25 as you do. To get the general formula, seperate the exposed faces into directions from which you can see them: top, bottom (only for question 2), left, right, front, behind. Some of those should be easy to calculate, others may need a step like for question 1, but simpler.
answered Nov 27 '18 at 23:23
IngixIngix
3,689146
$endgroup$
$begingroup$
You get 21 because you’re including the bottom face from when $n=1$ but Idk why u should include that since it isn’t “exposed”
$endgroup$
– user8290579
Nov 28 '18 at 0:33
$begingroup$
You are right, miscalulation on my part (you see $4times4$ faces from the sides, not $3times4$), so 25 is the correct values for $n=2$.
$endgroup$
– Ingix
Nov 28 '18 at 9:40
add a comment |
$begingroup$
For the first question how about considering the sequence, $U_{n}=(2n-1)^2 +U_{n-1}, forall nge 1$ and take $U_0=0$
With $U_n$ representing the number of cubes found in the pyramid up to level $n$
answered Nov 28 '18 at 16:36
Fareed AFFareed AF
45211
$endgroup$
$begingroup$
yeah I realized my mistake, it should be $1^2 +2^2 +...+ (2n-1)^2$ instead of cubed
$endgroup$
– user8290579
Nov 28 '18 at 16:43
add a comment |
$begingroup$
Hint:
By counting and generalizing the pattern,
The number of cubes is $1^2+3^2+5^2+7^2+cdots (2n-1)^2$.
The number of exposed faces, bottom excluded, is $1+4+20+36+52+cdots 8(2n-1)-4$.
There are $(2n-1)^2$ faces at the bottom.
You can rewrite the general terms as functions of $n$ and $n^2$ and use the Faulhaber formulas.
answered Nov 28 '18 at 16:55
Yves DaoustYves Daoust
126k671223
$endgroup$
add a comment |
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4 Answers
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4 Answers
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$begingroup$
It's hard to know what you are counting and what you are not counting.
You have three different questions; one can see when you stop discussing the first one and start on the other two, but you do not clearly state when you are working on part 2 and when on part 3.
Your use of the formula $4(2n-1)$ is confusing, because in this formula you decided to ignore the meaning of $n$ as given in the problem statement,
where $n$ is the number of levels
(which can have values $1, 2, 3, 4, ldots$).
Coincidentally, $4(2n-1)$ happens to be the number of faces on the $n$th level that are exposed on the vertical sides, not including the faces exposed on the top or bottom of any cube.
Instead of staying consistent with the notation in the problem statement,
you decided to use the variable $n$ to represent something different,
a number that can only take odd values.
You get lucky later because by the time you are trying to derive a formula
for the "real" value of $n$ as defined in the problem statement,
you have only explicit integers like $113$ which have "forgotten" your other definition of $n.$
An approach that would be less confusing for readers who aren't inside your head would be to introduce a new variable for the odd numbers you wanted to represent, for example $m.$
Then you are free to say $m$ has only odd values without contradicting anything in the problem statement, and you can write $4(2m-1)$ for the number of new faces added by the last level, where $m$ is the number of cubes along one edge of that level, which is always an odd number.
Note that if you define $m$ this way, then $m = 2n - 1$ where $n$ is the number of levels added so far.
At the end you got the correct sequence for part 3.
One may guess that you did this by taking partial sums of the series
$5 + 20 + 36 + 36 + 52 + 68 + cdots.$
It would be better if you said you were doing that.
After that point, although you did not provide all the details of the finite difference method, at least you said you were using finite differences so we know what you were trying to do there.
The difference between questions 2 and 3 has to do with the the implicit assumption that we are always dealing with a pyramid that is finite in size
(though it can be arbitrarily large, not limited to $11$ levels like the pyramid in the figure)
and also with the implicit assumption that
"the entire pyramid up to level $n$"
is a pyramid of exactly $n$ levels that you can look at from all angles, so that your ability to look at the bottom of the $n$th level of the pyramid is not blocked by being stuck to some even larger level underneath it.
To be fair, I think the problem would be much better worded if it explicitly said "a pyramid of $n$ levels."
If it were really a problem about the top $n$ levels of the exact pyramid shown in the figure, which has $11$ levels, then the formula for the number of exposed faces would be identical for questions 2 and 3 when $n < 11$ but would suddenly become much larger for question 2 when $n = 11.$
Also, the answers for $n > 11$ would all be numerically the same as for
$n = 11$; that is, since there are only $11$ levels, we don't find any more sides by attempting to count more than $11$ levels.
This would make a very silly problem, which is why I did not even consider this interpretation at first but only considered the interpretation where $n$ is the total number of levels in an arbitrary pyramid.
As an alternative approach, which I think is easier (although not useful as an exercise in using the method of finite differences),
you could look at the pyramid's orthogonal views from each of the six directions: above, below, left, right, front, back.
In each direction you see a set of exposed faces of cubes--a different set of faces in each direction, and all exposed faces are seen in one of the directions,
so all you have to do is find the number of faces seen from each direction
and add up those six numbers
(or in question 3, five numbers, since we do not include the number of faces on the bottom).
To make it even easier, the top and bottom views are identical large squares
of edge length $2n - 1$,
and the other four views are all arrangements of the form
$1 + 3 + 5 + cdots + (2n - 1)$ for $n$ layers,
which you should probably recognize and know how to express as a polynomial.
Using this approach one can quickly confirm that the formula you found,
$8n^2 - 4n + 1,$ is the correct answer for question 3.
We have $(2n - 1)^2 = 4n^2 - 4 + 1$ faces visible from above,
and $n^2$ faces visible from each of the other four views
(excluding the bottom view).
answered Nov 28 '18 at 4:42
David KDavid K
53.9k342116
$endgroup$
$begingroup$
The way that I got 4(2n-1) ad my formula is because my values for n are all odd numbers. For example, $n=3: 4(5)=20$ which is the correct value for the number of exposed faces for the 3 x 3. I explained how I got the answer for question 3, I created a sequence of numbers which represent the sum of all the exposed faces, and then did the finite differences method to find a polynomial which matches it.
$endgroup$
– user8290579
Nov 28 '18 at 14:30
$begingroup$
thank you very much for explaining the $4(2m-1), m = 2n-1$ I was really confused about that but this makes it a lot clearer and it would make it a lot clearer to other people. unfortunately the only method i can use is the finite differences method. and for the distinction between problems 2 and 3, you are basically saying that for problem 2, including the bottom faces means excluding the bottom faces for (n-1) levels and including it for the nth level, right? and for question 3, we are excluding the bottom faces for all n levels, right? Why am i getting correct answers w the formulas i have
$endgroup$
– user8290579
Nov 28 '18 at 16:42
$begingroup$
You are right, in question 2 the bottom faces are counted only on the $n$th level. It occurred to me (after writing the initial answer) that this might be intended as an exercise in finite differences, so solving it the "easy" way would get you no credit.
$endgroup$
– David K
Nov 28 '18 at 18:07
$begingroup$
Regarding the confusion over $4(n-1),$ notice that if you changed $n$ to $m$ just in the two paragraphs that mention $4(n-1),$ you don't have to change anything else to make sense of it. I think you just had (in your mind) a temporary definition of $n$ for those two paragraphs, and you used it to get the sequence $5,25,61,113,181,265,$ which is a correct sequence for part 3. Notice there is no $n$ in that sequence; it doesn't use the "temporary" definition of $n$ and you never go back to that definition. The next time $n$ appears you're using it exactly as intended in the problem statement.
$endgroup$
– David K
Nov 28 '18 at 18:13
1
$begingroup$
Got it, so if I add $4n^2-4n+1$ to the sequence $8n^2-4n+1$ then that would be the correct answer?
$endgroup$
– user8290579
Nov 28 '18 at 23:22
|
show 2 more comments
$begingroup$
It's hard to know what you are counting and what you are not counting.
You have three different questions; one can see when you stop discussing the first one and start on the other two, but you do not clearly state when you are working on part 2 and when on part 3.
Your use of the formula $4(2n-1)$ is confusing, because in this formula you decided to ignore the meaning of $n$ as given in the problem statement,
where $n$ is the number of levels
(which can have values $1, 2, 3, 4, ldots$).
Coincidentally, $4(2n-1)$ happens to be the number of faces on the $n$th level that are exposed on the vertical sides, not including the faces exposed on the top or bottom of any cube.
Instead of staying consistent with the notation in the problem statement,
you decided to use the variable $n$ to represent something different,
a number that can only take odd values.
You get lucky later because by the time you are trying to derive a formula
for the "real" value of $n$ as defined in the problem statement,
you have only explicit integers like $113$ which have "forgotten" your other definition of $n.$
An approach that would be less confusing for readers who aren't inside your head would be to introduce a new variable for the odd numbers you wanted to represent, for example $m.$
Then you are free to say $m$ has only odd values without contradicting anything in the problem statement, and you can write $4(2m-1)$ for the number of new faces added by the last level, where $m$ is the number of cubes along one edge of that level, which is always an odd number.
Note that if you define $m$ this way, then $m = 2n - 1$ where $n$ is the number of levels added so far.
At the end you got the correct sequence for part 3.
One may guess that you did this by taking partial sums of the series
$5 + 20 + 36 + 36 + 52 + 68 + cdots.$
It would be better if you said you were doing that.
After that point, although you did not provide all the details of the finite difference method, at least you said you were using finite differences so we know what you were trying to do there.
The difference between questions 2 and 3 has to do with the the implicit assumption that we are always dealing with a pyramid that is finite in size
(though it can be arbitrarily large, not limited to $11$ levels like the pyramid in the figure)
and also with the implicit assumption that
"the entire pyramid up to level $n$"
is a pyramid of exactly $n$ levels that you can look at from all angles, so that your ability to look at the bottom of the $n$th level of the pyramid is not blocked by being stuck to some even larger level underneath it.
To be fair, I think the problem would be much better worded if it explicitly said "a pyramid of $n$ levels."
If it were really a problem about the top $n$ levels of the exact pyramid shown in the figure, which has $11$ levels, then the formula for the number of exposed faces would be identical for questions 2 and 3 when $n < 11$ but would suddenly become much larger for question 2 when $n = 11.$
Also, the answers for $n > 11$ would all be numerically the same as for
$n = 11$; that is, since there are only $11$ levels, we don't find any more sides by attempting to count more than $11$ levels.
This would make a very silly problem, which is why I did not even consider this interpretation at first but only considered the interpretation where $n$ is the total number of levels in an arbitrary pyramid.
As an alternative approach, which I think is easier (although not useful as an exercise in using the method of finite differences),
you could look at the pyramid's orthogonal views from each of the six directions: above, below, left, right, front, back.
In each direction you see a set of exposed faces of cubes--a different set of faces in each direction, and all exposed faces are seen in one of the directions,
so all you have to do is find the number of faces seen from each direction
and add up those six numbers
(or in question 3, five numbers, since we do not include the number of faces on the bottom).
To make it even easier, the top and bottom views are identical large squares
of edge length $2n - 1$,
and the other four views are all arrangements of the form
$1 + 3 + 5 + cdots + (2n - 1)$ for $n$ layers,
which you should probably recognize and know how to express as a polynomial.
Using this approach one can quickly confirm that the formula you found,
$8n^2 - 4n + 1,$ is the correct answer for question 3.
We have $(2n - 1)^2 = 4n^2 - 4 + 1$ faces visible from above,
and $n^2$ faces visible from each of the other four views
(excluding the bottom view).
answered Nov 28 '18 at 4:42
David KDavid K
53.9k342116
$endgroup$
$begingroup$
The way that I got 4(2n-1) ad my formula is because my values for n are all odd numbers. For example, $n=3: 4(5)=20$ which is the correct value for the number of exposed faces for the 3 x 3. I explained how I got the answer for question 3, I created a sequence of numbers which represent the sum of all the exposed faces, and then did the finite differences method to find a polynomial which matches it.
$endgroup$
– user8290579
Nov 28 '18 at 14:30
$begingroup$
thank you very much for explaining the $4(2m-1), m = 2n-1$ I was really confused about that but this makes it a lot clearer and it would make it a lot clearer to other people. unfortunately the only method i can use is the finite differences method. and for the distinction between problems 2 and 3, you are basically saying that for problem 2, including the bottom faces means excluding the bottom faces for (n-1) levels and including it for the nth level, right? and for question 3, we are excluding the bottom faces for all n levels, right? Why am i getting correct answers w the formulas i have
$endgroup$
– user8290579
Nov 28 '18 at 16:42
$begingroup$
You are right, in question 2 the bottom faces are counted only on the $n$th level. It occurred to me (after writing the initial answer) that this might be intended as an exercise in finite differences, so solving it the "easy" way would get you no credit.
$endgroup$
– David K
Nov 28 '18 at 18:07
$begingroup$
Regarding the confusion over $4(n-1),$ notice that if you changed $n$ to $m$ just in the two paragraphs that mention $4(n-1),$ you don't have to change anything else to make sense of it. I think you just had (in your mind) a temporary definition of $n$ for those two paragraphs, and you used it to get the sequence $5,25,61,113,181,265,$ which is a correct sequence for part 3. Notice there is no $n$ in that sequence; it doesn't use the "temporary" definition of $n$ and you never go back to that definition. The next time $n$ appears you're using it exactly as intended in the problem statement.
$endgroup$
– David K
Nov 28 '18 at 18:13
1
$begingroup$
Got it, so if I add $4n^2-4n+1$ to the sequence $8n^2-4n+1$ then that would be the correct answer?
$endgroup$
– user8290579
Nov 28 '18 at 23:22
|
show 2 more comments
$begingroup$
It's hard to know what you are counting and what you are not counting.
You have three different questions; one can see when you stop discussing the first one and start on the other two, but you do not clearly state when you are working on part 2 and when on part 3.
Your use of the formula $4(2n-1)$ is confusing, because in this formula you decided to ignore the meaning of $n$ as given in the problem statement,
where $n$ is the number of levels
(which can have values $1, 2, 3, 4, ldots$).
Coincidentally, $4(2n-1)$ happens to be the number of faces on the $n$th level that are exposed on the vertical sides, not including the faces exposed on the top or bottom of any cube.
Instead of staying consistent with the notation in the problem statement,
you decided to use the variable $n$ to represent something different,
a number that can only take odd values.
You get lucky later because by the time you are trying to derive a formula
for the "real" value of $n$ as defined in the problem statement,
you have only explicit integers like $113$ which have "forgotten" your other definition of $n.$
An approach that would be less confusing for readers who aren't inside your head would be to introduce a new variable for the odd numbers you wanted to represent, for example $m.$
Then you are free to say $m$ has only odd values without contradicting anything in the problem statement, and you can write $4(2m-1)$ for the number of new faces added by the last level, where $m$ is the number of cubes along one edge of that level, which is always an odd number.
Note that if you define $m$ this way, then $m = 2n - 1$ where $n$ is the number of levels added so far.
At the end you got the correct sequence for part 3.
One may guess that you did this by taking partial sums of the series
$5 + 20 + 36 + 36 + 52 + 68 + cdots.$
It would be better if you said you were doing that.
After that point, although you did not provide all the details of the finite difference method, at least you said you were using finite differences so we know what you were trying to do there.
The difference between questions 2 and 3 has to do with the the implicit assumption that we are always dealing with a pyramid that is finite in size
(though it can be arbitrarily large, not limited to $11$ levels like the pyramid in the figure)
and also with the implicit assumption that
"the entire pyramid up to level $n$"
is a pyramid of exactly $n$ levels that you can look at from all angles, so that your ability to look at the bottom of the $n$th level of the pyramid is not blocked by being stuck to some even larger level underneath it.
To be fair, I think the problem would be much better worded if it explicitly said "a pyramid of $n$ levels."
If it were really a problem about the top $n$ levels of the exact pyramid shown in the figure, which has $11$ levels, then the formula for the number of exposed faces would be identical for questions 2 and 3 when $n < 11$ but would suddenly become much larger for question 2 when $n = 11.$
Also, the answers for $n > 11$ would all be numerically the same as for
$n = 11$; that is, since there are only $11$ levels, we don't find any more sides by attempting to count more than $11$ levels.
This would make a very silly problem, which is why I did not even consider this interpretation at first but only considered the interpretation where $n$ is the total number of levels in an arbitrary pyramid.
As an alternative approach, which I think is easier (although not useful as an exercise in using the method of finite differences),
you could look at the pyramid's orthogonal views from each of the six directions: above, below, left, right, front, back.
In each direction you see a set of exposed faces of cubes--a different set of faces in each direction, and all exposed faces are seen in one of the directions,
so all you have to do is find the number of faces seen from each direction
and add up those six numbers
(or in question 3, five numbers, since we do not include the number of faces on the bottom).
To make it even easier, the top and bottom views are identical large squares
of edge length $2n - 1$,
and the other four views are all arrangements of the form
$1 + 3 + 5 + cdots + (2n - 1)$ for $n$ layers,
which you should probably recognize and know how to express as a polynomial.
Using this approach one can quickly confirm that the formula you found,
$8n^2 - 4n + 1,$ is the correct answer for question 3.
We have $(2n - 1)^2 = 4n^2 - 4 + 1$ faces visible from above,
and $n^2$ faces visible from each of the other four views
(excluding the bottom view).
answered Nov 28 '18 at 4:42
David KDavid K
53.9k342116
$endgroup$
It's hard to know what you are counting and what you are not counting.
You have three different questions; one can see when you stop discussing the first one and start on the other two, but you do not clearly state when you are working on part 2 and when on part 3.
Your use of the formula $4(2n-1)$ is confusing, because in this formula you decided to ignore the meaning of $n$ as given in the problem statement,
where $n$ is the number of levels
(which can have values $1, 2, 3, 4, ldots$).
Coincidentally, $4(2n-1)$ happens to be the number of faces on the $n$th level that are exposed on the vertical sides, not including the faces exposed on the top or bottom of any cube.
Instead of staying consistent with the notation in the problem statement,
you decided to use the variable $n$ to represent something different,
a number that can only take odd values.
You get lucky later because by the time you are trying to derive a formula
for the "real" value of $n$ as defined in the problem statement,
you have only explicit integers like $113$ which have "forgotten" your other definition of $n.$
An approach that would be less confusing for readers who aren't inside your head would be to introduce a new variable for the odd numbers you wanted to represent, for example $m.$
Then you are free to say $m$ has only odd values without contradicting anything in the problem statement, and you can write $4(2m-1)$ for the number of new faces added by the last level, where $m$ is the number of cubes along one edge of that level, which is always an odd number.
Note that if you define $m$ this way, then $m = 2n - 1$ where $n$ is the number of levels added so far.
At the end you got the correct sequence for part 3.
One may guess that you did this by taking partial sums of the series
$5 + 20 + 36 + 36 + 52 + 68 + cdots.$
It would be better if you said you were doing that.
After that point, although you did not provide all the details of the finite difference method, at least you said you were using finite differences so we know what you were trying to do there.
The difference between questions 2 and 3 has to do with the the implicit assumption that we are always dealing with a pyramid that is finite in size
(though it can be arbitrarily large, not limited to $11$ levels like the pyramid in the figure)
and also with the implicit assumption that
"the entire pyramid up to level $n$"
is a pyramid of exactly $n$ levels that you can look at from all angles, so that your ability to look at the bottom of the $n$th level of the pyramid is not blocked by being stuck to some even larger level underneath it.
To be fair, I think the problem would be much better worded if it explicitly said "a pyramid of $n$ levels."
If it were really a problem about the top $n$ levels of the exact pyramid shown in the figure, which has $11$ levels, then the formula for the number of exposed faces would be identical for questions 2 and 3 when $n < 11$ but would suddenly become much larger for question 2 when $n = 11.$
Also, the answers for $n > 11$ would all be numerically the same as for
$n = 11$; that is, since there are only $11$ levels, we don't find any more sides by attempting to count more than $11$ levels.
This would make a very silly problem, which is why I did not even consider this interpretation at first but only considered the interpretation where $n$ is the total number of levels in an arbitrary pyramid.
As an alternative approach, which I think is easier (although not useful as an exercise in using the method of finite differences),
you could look at the pyramid's orthogonal views from each of the six directions: above, below, left, right, front, back.
In each direction you see a set of exposed faces of cubes--a different set of faces in each direction, and all exposed faces are seen in one of the directions,
so all you have to do is find the number of faces seen from each direction
and add up those six numbers
(or in question 3, five numbers, since we do not include the number of faces on the bottom).
To make it even easier, the top and bottom views are identical large squares
of edge length $2n - 1$,
and the other four views are all arrangements of the form
$1 + 3 + 5 + cdots + (2n - 1)$ for $n$ layers,
which you should probably recognize and know how to express as a polynomial.
Using this approach one can quickly confirm that the formula you found,
$8n^2 - 4n + 1,$ is the correct answer for question 3.
We have $(2n - 1)^2 = 4n^2 - 4 + 1$ faces visible from above,
and $n^2$ faces visible from each of the other four views
(excluding the bottom view).
answered Nov 28 '18 at 4:42
David KDavid K
53.9k342116
edited Nov 28 '18 at 16:19
answered Nov 28 '18 at 4:42
David KDavid K
53.9k342116
answered Nov 28 '18 at 4:42
David KDavid K
53.9k342116
answered Nov 28 '18 at 4:42
David KDavid K
53.9k342116
53.9k342116
$begingroup$
The way that I got 4(2n-1) ad my formula is because my values for n are all odd numbers. For example, $n=3: 4(5)=20$ which is the correct value for the number of exposed faces for the 3 x 3. I explained how I got the answer for question 3, I created a sequence of numbers which represent the sum of all the exposed faces, and then did the finite differences method to find a polynomial which matches it.
$endgroup$
– user8290579
Nov 28 '18 at 14:30
$begingroup$
thank you very much for explaining the $4(2m-1), m = 2n-1$ I was really confused about that but this makes it a lot clearer and it would make it a lot clearer to other people. unfortunately the only method i can use is the finite differences method. and for the distinction between problems 2 and 3, you are basically saying that for problem 2, including the bottom faces means excluding the bottom faces for (n-1) levels and including it for the nth level, right? and for question 3, we are excluding the bottom faces for all n levels, right? Why am i getting correct answers w the formulas i have
$endgroup$
– user8290579
Nov 28 '18 at 16:42
$begingroup$
You are right, in question 2 the bottom faces are counted only on the $n$th level. It occurred to me (after writing the initial answer) that this might be intended as an exercise in finite differences, so solving it the "easy" way would get you no credit.
$endgroup$
– David K
Nov 28 '18 at 18:07
$begingroup$
Regarding the confusion over $4(n-1),$ notice that if you changed $n$ to $m$ just in the two paragraphs that mention $4(n-1),$ you don't have to change anything else to make sense of it. I think you just had (in your mind) a temporary definition of $n$ for those two paragraphs, and you used it to get the sequence $5,25,61,113,181,265,$ which is a correct sequence for part 3. Notice there is no $n$ in that sequence; it doesn't use the "temporary" definition of $n$ and you never go back to that definition. The next time $n$ appears you're using it exactly as intended in the problem statement.
$endgroup$
– David K
Nov 28 '18 at 18:13
1
$begingroup$
Got it, so if I add $4n^2-4n+1$ to the sequence $8n^2-4n+1$ then that would be the correct answer?
$endgroup$
– user8290579
Nov 28 '18 at 23:22
|
show 2 more comments
$begingroup$
The way that I got 4(2n-1) ad my formula is because my values for n are all odd numbers. For example, $n=3: 4(5)=20$ which is the correct value for the number of exposed faces for the 3 x 3. I explained how I got the answer for question 3, I created a sequence of numbers which represent the sum of all the exposed faces, and then did the finite differences method to find a polynomial which matches it.
$endgroup$
– user8290579
Nov 28 '18 at 14:30
$begingroup$
thank you very much for explaining the $4(2m-1), m = 2n-1$ I was really confused about that but this makes it a lot clearer and it would make it a lot clearer to other people. unfortunately the only method i can use is the finite differences method. and for the distinction between problems 2 and 3, you are basically saying that for problem 2, including the bottom faces means excluding the bottom faces for (n-1) levels and including it for the nth level, right? and for question 3, we are excluding the bottom faces for all n levels, right? Why am i getting correct answers w the formulas i have
$endgroup$
– user8290579
Nov 28 '18 at 16:42
$begingroup$
You are right, in question 2 the bottom faces are counted only on the $n$th level. It occurred to me (after writing the initial answer) that this might be intended as an exercise in finite differences, so solving it the "easy" way would get you no credit.
$endgroup$
– David K
Nov 28 '18 at 18:07
$begingroup$
Regarding the confusion over $4(n-1),$ notice that if you changed $n$ to $m$ just in the two paragraphs that mention $4(n-1),$ you don't have to change anything else to make sense of it. I think you just had (in your mind) a temporary definition of $n$ for those two paragraphs, and you used it to get the sequence $5,25,61,113,181,265,$ which is a correct sequence for part 3. Notice there is no $n$ in that sequence; it doesn't use the "temporary" definition of $n$ and you never go back to that definition. The next time $n$ appears you're using it exactly as intended in the problem statement.
$endgroup$
– David K
Nov 28 '18 at 18:13
1
$begingroup$
Got it, so if I add $4n^2-4n+1$ to the sequence $8n^2-4n+1$ then that would be the correct answer?
$endgroup$
– user8290579
Nov 28 '18 at 23:22
$begingroup$
The way that I got 4(2n-1) ad my formula is because my values for n are all odd numbers. For example, $n=3: 4(5)=20$ which is the correct value for the number of exposed faces for the 3 x 3. I explained how I got the answer for question 3, I created a sequence of numbers which represent the sum of all the exposed faces, and then did the finite differences method to find a polynomial which matches it.
$endgroup$
– user8290579
Nov 28 '18 at 14:30
$begingroup$
The way that I got 4(2n-1) ad my formula is because my values for n are all odd numbers. For example, $n=3: 4(5)=20$ which is the correct value for the number of exposed faces for the 3 x 3. I explained how I got the answer for question 3, I created a sequence of numbers which represent the sum of all the exposed faces, and then did the finite differences method to find a polynomial which matches it.
$endgroup$
– user8290579
Nov 28 '18 at 14:30
$begingroup$
thank you very much for explaining the $4(2m-1), m = 2n-1$ I was really confused about that but this makes it a lot clearer and it would make it a lot clearer to other people. unfortunately the only method i can use is the finite differences method. and for the distinction between problems 2 and 3, you are basically saying that for problem 2, including the bottom faces means excluding the bottom faces for (n-1) levels and including it for the nth level, right? and for question 3, we are excluding the bottom faces for all n levels, right? Why am i getting correct answers w the formulas i have
$endgroup$
– user8290579
Nov 28 '18 at 16:42
$begingroup$
thank you very much for explaining the $4(2m-1), m = 2n-1$ I was really confused about that but this makes it a lot clearer and it would make it a lot clearer to other people. unfortunately the only method i can use is the finite differences method. and for the distinction between problems 2 and 3, you are basically saying that for problem 2, including the bottom faces means excluding the bottom faces for (n-1) levels and including it for the nth level, right? and for question 3, we are excluding the bottom faces for all n levels, right? Why am i getting correct answers w the formulas i have
$endgroup$
– user8290579
Nov 28 '18 at 16:42
$begingroup$
You are right, in question 2 the bottom faces are counted only on the $n$th level. It occurred to me (after writing the initial answer) that this might be intended as an exercise in finite differences, so solving it the "easy" way would get you no credit.
$endgroup$
– David K
Nov 28 '18 at 18:07
$begingroup$
You are right, in question 2 the bottom faces are counted only on the $n$th level. It occurred to me (after writing the initial answer) that this might be intended as an exercise in finite differences, so solving it the "easy" way would get you no credit.
$endgroup$
– David K
Nov 28 '18 at 18:07
$begingroup$
Regarding the confusion over $4(n-1),$ notice that if you changed $n$ to $m$ just in the two paragraphs that mention $4(n-1),$ you don't have to change anything else to make sense of it. I think you just had (in your mind) a temporary definition of $n$ for those two paragraphs, and you used it to get the sequence $5,25,61,113,181,265,$ which is a correct sequence for part 3. Notice there is no $n$ in that sequence; it doesn't use the "temporary" definition of $n$ and you never go back to that definition. The next time $n$ appears you're using it exactly as intended in the problem statement.
$endgroup$
– David K
Nov 28 '18 at 18:13
$begingroup$
Regarding the confusion over $4(n-1),$ notice that if you changed $n$ to $m$ just in the two paragraphs that mention $4(n-1),$ you don't have to change anything else to make sense of it. I think you just had (in your mind) a temporary definition of $n$ for those two paragraphs, and you used it to get the sequence $5,25,61,113,181,265,$ which is a correct sequence for part 3. Notice there is no $n$ in that sequence; it doesn't use the "temporary" definition of $n$ and you never go back to that definition. The next time $n$ appears you're using it exactly as intended in the problem statement.
$endgroup$
– David K
Nov 28 '18 at 18:13
1
1
$begingroup$
Got it, so if I add $4n^2-4n+1$ to the sequence $8n^2-4n+1$ then that would be the correct answer?
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– user8290579
Nov 28 '18 at 23:22
$begingroup$
Got it, so if I add $4n^2-4n+1$ to the sequence $8n^2-4n+1$ then that would be the correct answer?
$endgroup$
– user8290579
Nov 28 '18 at 23:22
|
show 2 more comments
$begingroup$
It's a bit much to comment on everything, but
for question 1): You are summing cubes (3rd powers), but that is not correct. Note that the number of cubes in the second layer from the top is $9=3^2$, nor $27=3^3$ as you wrote. The same goes of course for the other layers.
For question 2), 3): Based on your answer for $n=1$, you seem to be considering question 3 (no bottom faces). But I get 21 as the value for $n=2$, not 20 or 25 as you do. To get the general formula, seperate the exposed faces into directions from which you can see them: top, bottom (only for question 2), left, right, front, behind. Some of those should be easy to calculate, others may need a step like for question 1, but simpler.
answered Nov 27 '18 at 23:23
IngixIngix
3,689146
$endgroup$
$begingroup$
You get 21 because you’re including the bottom face from when $n=1$ but Idk why u should include that since it isn’t “exposed”
$endgroup$
– user8290579
Nov 28 '18 at 0:33
$begingroup$
You are right, miscalulation on my part (you see $4times4$ faces from the sides, not $3times4$), so 25 is the correct values for $n=2$.
$endgroup$
– Ingix
Nov 28 '18 at 9:40
add a comment |
$begingroup$
It's a bit much to comment on everything, but
for question 1): You are summing cubes (3rd powers), but that is not correct. Note that the number of cubes in the second layer from the top is $9=3^2$, nor $27=3^3$ as you wrote. The same goes of course for the other layers.
For question 2), 3): Based on your answer for $n=1$, you seem to be considering question 3 (no bottom faces). But I get 21 as the value for $n=2$, not 20 or 25 as you do. To get the general formula, seperate the exposed faces into directions from which you can see them: top, bottom (only for question 2), left, right, front, behind. Some of those should be easy to calculate, others may need a step like for question 1, but simpler.
answered Nov 27 '18 at 23:23
IngixIngix
3,689146
$endgroup$
$begingroup$
You get 21 because you’re including the bottom face from when $n=1$ but Idk why u should include that since it isn’t “exposed”
$endgroup$
– user8290579
Nov 28 '18 at 0:33
$begingroup$
You are right, miscalulation on my part (you see $4times4$ faces from the sides, not $3times4$), so 25 is the correct values for $n=2$.
$endgroup$
– Ingix
Nov 28 '18 at 9:40
add a comment |
$begingroup$
It's a bit much to comment on everything, but
for question 1): You are summing cubes (3rd powers), but that is not correct. Note that the number of cubes in the second layer from the top is $9=3^2$, nor $27=3^3$ as you wrote. The same goes of course for the other layers.
For question 2), 3): Based on your answer for $n=1$, you seem to be considering question 3 (no bottom faces). But I get 21 as the value for $n=2$, not 20 or 25 as you do. To get the general formula, seperate the exposed faces into directions from which you can see them: top, bottom (only for question 2), left, right, front, behind. Some of those should be easy to calculate, others may need a step like for question 1, but simpler.
answered Nov 27 '18 at 23:23
IngixIngix
3,689146
$endgroup$
It's a bit much to comment on everything, but
for question 1): You are summing cubes (3rd powers), but that is not correct. Note that the number of cubes in the second layer from the top is $9=3^2$, nor $27=3^3$ as you wrote. The same goes of course for the other layers.
For question 2), 3): Based on your answer for $n=1$, you seem to be considering question 3 (no bottom faces). But I get 21 as the value for $n=2$, not 20 or 25 as you do. To get the general formula, seperate the exposed faces into directions from which you can see them: top, bottom (only for question 2), left, right, front, behind. Some of those should be easy to calculate, others may need a step like for question 1, but simpler.
answered Nov 27 '18 at 23:23
IngixIngix
3,689146
answered Nov 27 '18 at 23:23
IngixIngix
3,689146
answered Nov 27 '18 at 23:23
IngixIngix
3,689146
answered Nov 27 '18 at 23:23
IngixIngix
3,689146
3,689146
$begingroup$
You get 21 because you’re including the bottom face from when $n=1$ but Idk why u should include that since it isn’t “exposed”
$endgroup$
– user8290579
Nov 28 '18 at 0:33
$begingroup$
You are right, miscalulation on my part (you see $4times4$ faces from the sides, not $3times4$), so 25 is the correct values for $n=2$.
$endgroup$
– Ingix
Nov 28 '18 at 9:40
add a comment |
$begingroup$
You get 21 because you’re including the bottom face from when $n=1$ but Idk why u should include that since it isn’t “exposed”
$endgroup$
– user8290579
Nov 28 '18 at 0:33
$begingroup$
You are right, miscalulation on my part (you see $4times4$ faces from the sides, not $3times4$), so 25 is the correct values for $n=2$.
$endgroup$
– Ingix
Nov 28 '18 at 9:40
$begingroup$
You get 21 because you’re including the bottom face from when $n=1$ but Idk why u should include that since it isn’t “exposed”
$endgroup$
– user8290579
Nov 28 '18 at 0:33
$begingroup$
You get 21 because you’re including the bottom face from when $n=1$ but Idk why u should include that since it isn’t “exposed”
$endgroup$
– user8290579
Nov 28 '18 at 0:33
$begingroup$
You are right, miscalulation on my part (you see $4times4$ faces from the sides, not $3times4$), so 25 is the correct values for $n=2$.
$endgroup$
– Ingix
Nov 28 '18 at 9:40
$begingroup$
You are right, miscalulation on my part (you see $4times4$ faces from the sides, not $3times4$), so 25 is the correct values for $n=2$.
$endgroup$
– Ingix
Nov 28 '18 at 9:40
add a comment |
$begingroup$
For the first question how about considering the sequence, $U_{n}=(2n-1)^2 +U_{n-1}, forall nge 1$ and take $U_0=0$
With $U_n$ representing the number of cubes found in the pyramid up to level $n$
answered Nov 28 '18 at 16:36
Fareed AFFareed AF
45211
$endgroup$
$begingroup$
yeah I realized my mistake, it should be $1^2 +2^2 +...+ (2n-1)^2$ instead of cubed
$endgroup$
– user8290579
Nov 28 '18 at 16:43
add a comment |
$begingroup$
For the first question how about considering the sequence, $U_{n}=(2n-1)^2 +U_{n-1}, forall nge 1$ and take $U_0=0$
With $U_n$ representing the number of cubes found in the pyramid up to level $n$
answered Nov 28 '18 at 16:36
Fareed AFFareed AF
45211
$endgroup$
$begingroup$
yeah I realized my mistake, it should be $1^2 +2^2 +...+ (2n-1)^2$ instead of cubed
$endgroup$
– user8290579
Nov 28 '18 at 16:43
add a comment |
$begingroup$
For the first question how about considering the sequence, $U_{n}=(2n-1)^2 +U_{n-1}, forall nge 1$ and take $U_0=0$
With $U_n$ representing the number of cubes found in the pyramid up to level $n$
answered Nov 28 '18 at 16:36
Fareed AFFareed AF
45211
$endgroup$
For the first question how about considering the sequence, $U_{n}=(2n-1)^2 +U_{n-1}, forall nge 1$ and take $U_0=0$
With $U_n$ representing the number of cubes found in the pyramid up to level $n$
answered Nov 28 '18 at 16:36
Fareed AFFareed AF
45211
edited Nov 28 '18 at 16:41
answered Nov 28 '18 at 16:36
Fareed AFFareed AF
45211
answered Nov 28 '18 at 16:36
Fareed AFFareed AF
45211
answered Nov 28 '18 at 16:36
Fareed AFFareed AF
45211
45211
$begingroup$
yeah I realized my mistake, it should be $1^2 +2^2 +...+ (2n-1)^2$ instead of cubed
$endgroup$
– user8290579
Nov 28 '18 at 16:43
add a comment |
$begingroup$
yeah I realized my mistake, it should be $1^2 +2^2 +...+ (2n-1)^2$ instead of cubed
$endgroup$
– user8290579
Nov 28 '18 at 16:43
$begingroup$
yeah I realized my mistake, it should be $1^2 +2^2 +...+ (2n-1)^2$ instead of cubed
$endgroup$
– user8290579
Nov 28 '18 at 16:43
$begingroup$
yeah I realized my mistake, it should be $1^2 +2^2 +...+ (2n-1)^2$ instead of cubed
$endgroup$
– user8290579
Nov 28 '18 at 16:43
add a comment |
$begingroup$
Hint:
By counting and generalizing the pattern,
The number of cubes is $1^2+3^2+5^2+7^2+cdots (2n-1)^2$.
The number of exposed faces, bottom excluded, is $1+4+20+36+52+cdots 8(2n-1)-4$.
There are $(2n-1)^2$ faces at the bottom.
You can rewrite the general terms as functions of $n$ and $n^2$ and use the Faulhaber formulas.
answered Nov 28 '18 at 16:55
Yves DaoustYves Daoust
126k671223
$endgroup$
add a comment |
$begingroup$
Hint:
By counting and generalizing the pattern,
The number of cubes is $1^2+3^2+5^2+7^2+cdots (2n-1)^2$.
The number of exposed faces, bottom excluded, is $1+4+20+36+52+cdots 8(2n-1)-4$.
There are $(2n-1)^2$ faces at the bottom.
You can rewrite the general terms as functions of $n$ and $n^2$ and use the Faulhaber formulas.
answered Nov 28 '18 at 16:55
Yves DaoustYves Daoust
126k671223
$endgroup$
add a comment |
$begingroup$
Hint:
By counting and generalizing the pattern,
The number of cubes is $1^2+3^2+5^2+7^2+cdots (2n-1)^2$.
The number of exposed faces, bottom excluded, is $1+4+20+36+52+cdots 8(2n-1)-4$.
There are $(2n-1)^2$ faces at the bottom.
You can rewrite the general terms as functions of $n$ and $n^2$ and use the Faulhaber formulas.
answered Nov 28 '18 at 16:55
Yves DaoustYves Daoust
126k671223
$endgroup$
Hint:
By counting and generalizing the pattern,
The number of cubes is $1^2+3^2+5^2+7^2+cdots (2n-1)^2$.
The number of exposed faces, bottom excluded, is $1+4+20+36+52+cdots 8(2n-1)-4$.
There are $(2n-1)^2$ faces at the bottom.
You can rewrite the general terms as functions of $n$ and $n^2$ and use the Faulhaber formulas.
answered Nov 28 '18 at 16:55
Yves DaoustYves Daoust
126k671223
answered Nov 28 '18 at 16:55
Yves DaoustYves Daoust
126k671223
answered Nov 28 '18 at 16:55
Yves DaoustYves Daoust
126k671223
answered Nov 28 '18 at 16:55
Yves DaoustYves Daoust
126k671223
126k671223
add a comment |
add a comment |
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$begingroup$
Your answer for 3) is correct, for 2) you made the error of assuming the bottom layer is $ntimes n$, but it is $(2n-1)times(2n-1)$. Of course, the arguments you gave may convince you that the forumulas are correct, but the idea that the formula is some kind of polynom and you can find it via the difference scheme is not bad but also not convincing to somebody who has never seen such a problem. For 1) my comment from the answer below still stands: Summing 3rd powers is not the right idea.
$endgroup$
– Ingix
Nov 28 '18 at 9:52