Given that the coefficients of the $6^{th}$ term and $16^{th}$ term in the Expansion of $(x + y) ^n$ are...
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The Expansion gives me: $6^{th}$ term = $displaystylebinom{n}{5} x^{n-5}y^5$ and
$16^{th}$ term = $displaystylebinom{n}{15} x^{n-15}y^{15}$. What do I do next?
binomial-coefficients
edited Nov 26 '18 at 10:34
Tianlalu
3,08621038
asked Nov 26 '18 at 9:07
Bravie Bravie
143
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closed as off-topic by Did, Paul Frost, Davide Giraudo, Jyrki Lahtonen, Shaun Nov 26 '18 at 13:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Paul Frost, Davide Giraudo, Jyrki Lahtonen, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
$begingroup$
The Expansion gives me: $6^{th}$ term = $displaystylebinom{n}{5} x^{n-5}y^5$ and
$16^{th}$ term = $displaystylebinom{n}{15} x^{n-15}y^{15}$. What do I do next?
binomial-coefficients
edited Nov 26 '18 at 10:34
Tianlalu
3,08621038
asked Nov 26 '18 at 9:07
Bravie Bravie
143
$endgroup$
closed as off-topic by Did, Paul Frost, Davide Giraudo, Jyrki Lahtonen, Shaun Nov 26 '18 at 13:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Paul Frost, Davide Giraudo, Jyrki Lahtonen, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Please edit the question so that it is readable. what is (n 5)? is it nC5? What is (y) 15?
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– NoChance
Nov 26 '18 at 9:21
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i.stack.imgur.com/4jUOr.png
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– Did
Nov 26 '18 at 10:30
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@Did: Sir this is with atmost humility I am asking. If $n=0, n=1, n=2, n=3, n=4$ then what will the $6^{th}$ term and the $16^{th}$ term be if not $ 0$?
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– Yadati Kiran
Nov 26 '18 at 10:44
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@YadatiKiran Again, did I write anything resembling this assertion? Since I have not, why are you asking me this? IOW, what can the phrasethen n=0, n=1, n=2, n=3 and n=4 are solutions, taken from my comment above, possibly mean in your opinion? Sorry but posting random comments will not change the fact that your post below is not a complete answer.
$endgroup$
– Did
Nov 26 '18 at 10:46
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Assume one is looking for nonnegative integers, then n=0, n=1, n=2, n=3 and n=4 are solutions. The other solutions solve the identity Γ(a+1)Γ(n−a+1)=Γ(b+1)Γ(n−b+1) for (a,b)=(15,5) It appears that the a−b roots of this are distinct, equal to a+b−12+a+b+12e2iℓπ/(a−b) for ℓ∈{0,1,…,a−b−1}. If this holds, the only real roots are n=a+b and, if a+b is even, n=−1. So the only supplementary nonnegative integer root is n=a+b=20 (Note that none of this seems to be even mentioned [below].) – Did (copy sorry for the lack of format)
$endgroup$
– quid♦
Nov 26 '18 at 12:28
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show 1 more comment
$begingroup$
The Expansion gives me: $6^{th}$ term = $displaystylebinom{n}{5} x^{n-5}y^5$ and
$16^{th}$ term = $displaystylebinom{n}{15} x^{n-15}y^{15}$. What do I do next?
binomial-coefficients
edited Nov 26 '18 at 10:34
Tianlalu
3,08621038
asked Nov 26 '18 at 9:07
Bravie Bravie
143
$endgroup$
The Expansion gives me: $6^{th}$ term = $displaystylebinom{n}{5} x^{n-5}y^5$ and
$16^{th}$ term = $displaystylebinom{n}{15} x^{n-15}y^{15}$. What do I do next?
binomial-coefficients
binomial-coefficients
edited Nov 26 '18 at 10:34
Tianlalu
3,08621038
asked Nov 26 '18 at 9:07
Bravie Bravie
143
edited Nov 26 '18 at 10:34
Tianlalu
3,08621038
asked Nov 26 '18 at 9:07
Bravie Bravie
143
edited Nov 26 '18 at 10:34
Tianlalu
3,08621038
edited Nov 26 '18 at 10:34
Tianlalu
3,08621038
edited Nov 26 '18 at 10:34
Tianlalu
3,08621038
3,08621038
asked Nov 26 '18 at 9:07
Bravie Bravie
143
asked Nov 26 '18 at 9:07
Bravie Bravie
143
asked Nov 26 '18 at 9:07
Bravie Bravie
143
143
closed as off-topic by Did, Paul Frost, Davide Giraudo, Jyrki Lahtonen, Shaun Nov 26 '18 at 13:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Paul Frost, Davide Giraudo, Jyrki Lahtonen, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Paul Frost, Davide Giraudo, Jyrki Lahtonen, Shaun Nov 26 '18 at 13:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Paul Frost, Davide Giraudo, Jyrki Lahtonen, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Please edit the question so that it is readable. what is (n 5)? is it nC5? What is (y) 15?
$endgroup$
– NoChance
Nov 26 '18 at 9:21
$begingroup$
i.stack.imgur.com/4jUOr.png
$endgroup$
– Did
Nov 26 '18 at 10:30
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@Did: Sir this is with atmost humility I am asking. If $n=0, n=1, n=2, n=3, n=4$ then what will the $6^{th}$ term and the $16^{th}$ term be if not $ 0$?
$endgroup$
– Yadati Kiran
Nov 26 '18 at 10:44
$begingroup$
@YadatiKiran Again, did I write anything resembling this assertion? Since I have not, why are you asking me this? IOW, what can the phrasethen n=0, n=1, n=2, n=3 and n=4 are solutions, taken from my comment above, possibly mean in your opinion? Sorry but posting random comments will not change the fact that your post below is not a complete answer.
$endgroup$
– Did
Nov 26 '18 at 10:46
$begingroup$
Assume one is looking for nonnegative integers, then n=0, n=1, n=2, n=3 and n=4 are solutions. The other solutions solve the identity Γ(a+1)Γ(n−a+1)=Γ(b+1)Γ(n−b+1) for (a,b)=(15,5) It appears that the a−b roots of this are distinct, equal to a+b−12+a+b+12e2iℓπ/(a−b) for ℓ∈{0,1,…,a−b−1}. If this holds, the only real roots are n=a+b and, if a+b is even, n=−1. So the only supplementary nonnegative integer root is n=a+b=20 (Note that none of this seems to be even mentioned [below].) – Did (copy sorry for the lack of format)
$endgroup$
– quid♦
Nov 26 '18 at 12:28
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show 1 more comment
1
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Please edit the question so that it is readable. what is (n 5)? is it nC5? What is (y) 15?
$endgroup$
– NoChance
Nov 26 '18 at 9:21
$begingroup$
i.stack.imgur.com/4jUOr.png
$endgroup$
– Did
Nov 26 '18 at 10:30
$begingroup$
@Did: Sir this is with atmost humility I am asking. If $n=0, n=1, n=2, n=3, n=4$ then what will the $6^{th}$ term and the $16^{th}$ term be if not $ 0$?
$endgroup$
– Yadati Kiran
Nov 26 '18 at 10:44
$begingroup$
@YadatiKiran Again, did I write anything resembling this assertion? Since I have not, why are you asking me this? IOW, what can the phrasethen n=0, n=1, n=2, n=3 and n=4 are solutions, taken from my comment above, possibly mean in your opinion? Sorry but posting random comments will not change the fact that your post below is not a complete answer.
$endgroup$
– Did
Nov 26 '18 at 10:46
$begingroup$
Assume one is looking for nonnegative integers, then n=0, n=1, n=2, n=3 and n=4 are solutions. The other solutions solve the identity Γ(a+1)Γ(n−a+1)=Γ(b+1)Γ(n−b+1) for (a,b)=(15,5) It appears that the a−b roots of this are distinct, equal to a+b−12+a+b+12e2iℓπ/(a−b) for ℓ∈{0,1,…,a−b−1}. If this holds, the only real roots are n=a+b and, if a+b is even, n=−1. So the only supplementary nonnegative integer root is n=a+b=20 (Note that none of this seems to be even mentioned [below].) – Did (copy sorry for the lack of format)
$endgroup$
– quid♦
Nov 26 '18 at 12:28
1
1
$begingroup$
Please edit the question so that it is readable. what is (n 5)? is it nC5? What is (y) 15?
$endgroup$
– NoChance
Nov 26 '18 at 9:21
$begingroup$
Please edit the question so that it is readable. what is (n 5)? is it nC5? What is (y) 15?
$endgroup$
– NoChance
Nov 26 '18 at 9:21
$begingroup$
i.stack.imgur.com/4jUOr.png
$endgroup$
– Did
Nov 26 '18 at 10:30
$begingroup$
i.stack.imgur.com/4jUOr.png
$endgroup$
– Did
Nov 26 '18 at 10:30
$begingroup$
@Did: Sir this is with atmost humility I am asking. If $n=0, n=1, n=2, n=3, n=4$ then what will the $6^{th}$ term and the $16^{th}$ term be if not $ 0$?
$endgroup$
– Yadati Kiran
Nov 26 '18 at 10:44
$begingroup$
@Did: Sir this is with atmost humility I am asking. If $n=0, n=1, n=2, n=3, n=4$ then what will the $6^{th}$ term and the $16^{th}$ term be if not $ 0$?
$endgroup$
– Yadati Kiran
Nov 26 '18 at 10:44
$begingroup$
@YadatiKiran Again, did I write anything resembling this assertion? Since I have not, why are you asking me this? IOW, what can the phrase
then n=0, n=1, n=2, n=3 and n=4 are solutions, taken from my comment above, possibly mean in your opinion? Sorry but posting random comments will not change the fact that your post below is not a complete answer.$endgroup$
– Did
Nov 26 '18 at 10:46
$begingroup$
@YadatiKiran Again, did I write anything resembling this assertion? Since I have not, why are you asking me this? IOW, what can the phrase
then n=0, n=1, n=2, n=3 and n=4 are solutions, taken from my comment above, possibly mean in your opinion? Sorry but posting random comments will not change the fact that your post below is not a complete answer.$endgroup$
– Did
Nov 26 '18 at 10:46
$begingroup$
Assume one is looking for nonnegative integers, then n=0, n=1, n=2, n=3 and n=4 are solutions. The other solutions solve the identity Γ(a+1)Γ(n−a+1)=Γ(b+1)Γ(n−b+1) for (a,b)=(15,5) It appears that the a−b roots of this are distinct, equal to a+b−12+a+b+12e2iℓπ/(a−b) for ℓ∈{0,1,…,a−b−1}. If this holds, the only real roots are n=a+b and, if a+b is even, n=−1. So the only supplementary nonnegative integer root is n=a+b=20 (Note that none of this seems to be even mentioned [below].) – Did (copy sorry for the lack of format)
$endgroup$
– quid♦
Nov 26 '18 at 12:28
$begingroup$
Assume one is looking for nonnegative integers, then n=0, n=1, n=2, n=3 and n=4 are solutions. The other solutions solve the identity Γ(a+1)Γ(n−a+1)=Γ(b+1)Γ(n−b+1) for (a,b)=(15,5) It appears that the a−b roots of this are distinct, equal to a+b−12+a+b+12e2iℓπ/(a−b) for ℓ∈{0,1,…,a−b−1}. If this holds, the only real roots are n=a+b and, if a+b is even, n=−1. So the only supplementary nonnegative integer root is n=a+b=20 (Note that none of this seems to be even mentioned [below].) – Did (copy sorry for the lack of format)
$endgroup$
– quid♦
Nov 26 '18 at 12:28
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1 Answer
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$dfrac{n!}{5!(n-5)!}=dfrac{n!}{15!(n-15)!} $. So $n$ is $20$.
$rule{17cm}{0.5pt}$
For completeness $left(text{Courtesy:} : textbf{Did} ,: textbf{gandalf61} ,: textbf{bof} ,: textbf{Francis}right)$
$\text{(*I agree this explanation still has a lot of details missing)}$
The coefficient of $x^ky^{n−k}$ is equal to the coefficient of $x^{n−k}y^k$. So we have $n−5=15n−5=15$ and hence $ n=20$.
$underline{text{Other solutions to}:: displaystylebinom{n}{5}=binom{n}{15}quad(**)}$
The binomial coefficient is defined as $$displaystyle binom{n}{k}=begin{cases}dfrac{n!}{k!(n-k)!} &0leq k leq n \ 0 &text{otherwise} end{cases}$$ So by definition, $n=0,1,2,3,4$ are also solutions to $(**)$.
$$displaystyle binom{n}{5}=binom{n}{15}implies displaystyle dfrac{15!}{5!}=dfrac{(n-5)!}{(n-15)!}$$
If $zinmathbb{C}$ and $a,binmathbb{Z}:$ we have $$dfrac{Gamma(z-a+1)}{ Gamma(z-b+1)}=(-1)^{b-a}frac{Gamma(b-z)}{ Gamma(a-z)}$$ Here $a=5, b=15$.
$$dfrac{Gamma(z-5+1)}{ Gamma(z-15+1)}=(-1)^{15-5}frac{Gamma(15-z)}{ Gamma(5-z)}=prod_{i=5}^{14}(i-z)$$
So $displaystyle dfrac{15!}{5!}= prod_{i=5}^{14}(i-z)$ has $10$ roots by fundamental theorem of algebra which are complex (requires more details).
$rule{17cm}{0.5pt}$
So for our question here it is reasonable to assume $n$ to be positive, $ngeq k$ and the expansion of $(x+y)^n$ to have $n+1$ terms.
So $fbox{$n$=$20$}$
answered Nov 26 '18 at 9:18
Yadati KiranYadati Kiran
1,749619
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Yadati, how did you arrive at n=20.? Am a bit behind please. Usually, I have difficulties dealing with factorials of this nature
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– Bravie
Nov 26 '18 at 9:41
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@NoChance: The question does say the coefficients are equal.
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– Yadati Kiran
Nov 26 '18 at 9:45
1
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@bof: True, but then you need to expand the binomial coefficient to complex plane through Gamma function, and it can get quite complicated I think.
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– Francis
Nov 26 '18 at 9:46
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The coefficients in the binomial expansion are symmetric - the coefficient of $x^ky^{n-k}$ is equal to the coefficient of $x^{n-k}y^k$. So in this case we must have $n-5=15$ and hence $n=20$.
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– gandalf61
Nov 26 '18 at 9:52
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@gandalf61 You are repeating the argument showing that $n=20$ is a solution, but not advancing one iota in the direction of a proof that it is the only one. As a matter of fact, $n=20$ is not the only solution since $n=0$, $1$, $2$, $3$ and $4$ also are, in contradiction to your argument.
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– Did
Nov 26 '18 at 10:00
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$dfrac{n!}{5!(n-5)!}=dfrac{n!}{15!(n-15)!} $. So $n$ is $20$.
$rule{17cm}{0.5pt}$
For completeness $left(text{Courtesy:} : textbf{Did} ,: textbf{gandalf61} ,: textbf{bof} ,: textbf{Francis}right)$
$\text{(*I agree this explanation still has a lot of details missing)}$
The coefficient of $x^ky^{n−k}$ is equal to the coefficient of $x^{n−k}y^k$. So we have $n−5=15n−5=15$ and hence $ n=20$.
$underline{text{Other solutions to}:: displaystylebinom{n}{5}=binom{n}{15}quad(**)}$
The binomial coefficient is defined as $$displaystyle binom{n}{k}=begin{cases}dfrac{n!}{k!(n-k)!} &0leq k leq n \ 0 &text{otherwise} end{cases}$$ So by definition, $n=0,1,2,3,4$ are also solutions to $(**)$.
$$displaystyle binom{n}{5}=binom{n}{15}implies displaystyle dfrac{15!}{5!}=dfrac{(n-5)!}{(n-15)!}$$
If $zinmathbb{C}$ and $a,binmathbb{Z}:$ we have $$dfrac{Gamma(z-a+1)}{ Gamma(z-b+1)}=(-1)^{b-a}frac{Gamma(b-z)}{ Gamma(a-z)}$$ Here $a=5, b=15$.
$$dfrac{Gamma(z-5+1)}{ Gamma(z-15+1)}=(-1)^{15-5}frac{Gamma(15-z)}{ Gamma(5-z)}=prod_{i=5}^{14}(i-z)$$
So $displaystyle dfrac{15!}{5!}= prod_{i=5}^{14}(i-z)$ has $10$ roots by fundamental theorem of algebra which are complex (requires more details).
$rule{17cm}{0.5pt}$
So for our question here it is reasonable to assume $n$ to be positive, $ngeq k$ and the expansion of $(x+y)^n$ to have $n+1$ terms.
So $fbox{$n$=$20$}$
answered Nov 26 '18 at 9:18
Yadati KiranYadati Kiran
1,749619
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Yadati, how did you arrive at n=20.? Am a bit behind please. Usually, I have difficulties dealing with factorials of this nature
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– Bravie
Nov 26 '18 at 9:41
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@NoChance: The question does say the coefficients are equal.
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– Yadati Kiran
Nov 26 '18 at 9:45
1
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@bof: True, but then you need to expand the binomial coefficient to complex plane through Gamma function, and it can get quite complicated I think.
$endgroup$
– Francis
Nov 26 '18 at 9:46
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The coefficients in the binomial expansion are symmetric - the coefficient of $x^ky^{n-k}$ is equal to the coefficient of $x^{n-k}y^k$. So in this case we must have $n-5=15$ and hence $n=20$.
$endgroup$
– gandalf61
Nov 26 '18 at 9:52
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@gandalf61 You are repeating the argument showing that $n=20$ is a solution, but not advancing one iota in the direction of a proof that it is the only one. As a matter of fact, $n=20$ is not the only solution since $n=0$, $1$, $2$, $3$ and $4$ also are, in contradiction to your argument.
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– Did
Nov 26 '18 at 10:00
|
show 10 more comments
$begingroup$
$dfrac{n!}{5!(n-5)!}=dfrac{n!}{15!(n-15)!} $. So $n$ is $20$.
$rule{17cm}{0.5pt}$
For completeness $left(text{Courtesy:} : textbf{Did} ,: textbf{gandalf61} ,: textbf{bof} ,: textbf{Francis}right)$
$\text{(*I agree this explanation still has a lot of details missing)}$
The coefficient of $x^ky^{n−k}$ is equal to the coefficient of $x^{n−k}y^k$. So we have $n−5=15n−5=15$ and hence $ n=20$.
$underline{text{Other solutions to}:: displaystylebinom{n}{5}=binom{n}{15}quad(**)}$
The binomial coefficient is defined as $$displaystyle binom{n}{k}=begin{cases}dfrac{n!}{k!(n-k)!} &0leq k leq n \ 0 &text{otherwise} end{cases}$$ So by definition, $n=0,1,2,3,4$ are also solutions to $(**)$.
$$displaystyle binom{n}{5}=binom{n}{15}implies displaystyle dfrac{15!}{5!}=dfrac{(n-5)!}{(n-15)!}$$
If $zinmathbb{C}$ and $a,binmathbb{Z}:$ we have $$dfrac{Gamma(z-a+1)}{ Gamma(z-b+1)}=(-1)^{b-a}frac{Gamma(b-z)}{ Gamma(a-z)}$$ Here $a=5, b=15$.
$$dfrac{Gamma(z-5+1)}{ Gamma(z-15+1)}=(-1)^{15-5}frac{Gamma(15-z)}{ Gamma(5-z)}=prod_{i=5}^{14}(i-z)$$
So $displaystyle dfrac{15!}{5!}= prod_{i=5}^{14}(i-z)$ has $10$ roots by fundamental theorem of algebra which are complex (requires more details).
$rule{17cm}{0.5pt}$
So for our question here it is reasonable to assume $n$ to be positive, $ngeq k$ and the expansion of $(x+y)^n$ to have $n+1$ terms.
So $fbox{$n$=$20$}$
answered Nov 26 '18 at 9:18
Yadati KiranYadati Kiran
1,749619
$endgroup$
$begingroup$
Yadati, how did you arrive at n=20.? Am a bit behind please. Usually, I have difficulties dealing with factorials of this nature
$endgroup$
– Bravie
Nov 26 '18 at 9:41
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@NoChance: The question does say the coefficients are equal.
$endgroup$
– Yadati Kiran
Nov 26 '18 at 9:45
1
$begingroup$
@bof: True, but then you need to expand the binomial coefficient to complex plane through Gamma function, and it can get quite complicated I think.
$endgroup$
– Francis
Nov 26 '18 at 9:46
$begingroup$
The coefficients in the binomial expansion are symmetric - the coefficient of $x^ky^{n-k}$ is equal to the coefficient of $x^{n-k}y^k$. So in this case we must have $n-5=15$ and hence $n=20$.
$endgroup$
– gandalf61
Nov 26 '18 at 9:52
$begingroup$
@gandalf61 You are repeating the argument showing that $n=20$ is a solution, but not advancing one iota in the direction of a proof that it is the only one. As a matter of fact, $n=20$ is not the only solution since $n=0$, $1$, $2$, $3$ and $4$ also are, in contradiction to your argument.
$endgroup$
– Did
Nov 26 '18 at 10:00
|
show 10 more comments
$begingroup$
$dfrac{n!}{5!(n-5)!}=dfrac{n!}{15!(n-15)!} $. So $n$ is $20$.
$rule{17cm}{0.5pt}$
For completeness $left(text{Courtesy:} : textbf{Did} ,: textbf{gandalf61} ,: textbf{bof} ,: textbf{Francis}right)$
$\text{(*I agree this explanation still has a lot of details missing)}$
The coefficient of $x^ky^{n−k}$ is equal to the coefficient of $x^{n−k}y^k$. So we have $n−5=15n−5=15$ and hence $ n=20$.
$underline{text{Other solutions to}:: displaystylebinom{n}{5}=binom{n}{15}quad(**)}$
The binomial coefficient is defined as $$displaystyle binom{n}{k}=begin{cases}dfrac{n!}{k!(n-k)!} &0leq k leq n \ 0 &text{otherwise} end{cases}$$ So by definition, $n=0,1,2,3,4$ are also solutions to $(**)$.
$$displaystyle binom{n}{5}=binom{n}{15}implies displaystyle dfrac{15!}{5!}=dfrac{(n-5)!}{(n-15)!}$$
If $zinmathbb{C}$ and $a,binmathbb{Z}:$ we have $$dfrac{Gamma(z-a+1)}{ Gamma(z-b+1)}=(-1)^{b-a}frac{Gamma(b-z)}{ Gamma(a-z)}$$ Here $a=5, b=15$.
$$dfrac{Gamma(z-5+1)}{ Gamma(z-15+1)}=(-1)^{15-5}frac{Gamma(15-z)}{ Gamma(5-z)}=prod_{i=5}^{14}(i-z)$$
So $displaystyle dfrac{15!}{5!}= prod_{i=5}^{14}(i-z)$ has $10$ roots by fundamental theorem of algebra which are complex (requires more details).
$rule{17cm}{0.5pt}$
So for our question here it is reasonable to assume $n$ to be positive, $ngeq k$ and the expansion of $(x+y)^n$ to have $n+1$ terms.
So $fbox{$n$=$20$}$
answered Nov 26 '18 at 9:18
Yadati KiranYadati Kiran
1,749619
$endgroup$
$dfrac{n!}{5!(n-5)!}=dfrac{n!}{15!(n-15)!} $. So $n$ is $20$.
$rule{17cm}{0.5pt}$
For completeness $left(text{Courtesy:} : textbf{Did} ,: textbf{gandalf61} ,: textbf{bof} ,: textbf{Francis}right)$
$\text{(*I agree this explanation still has a lot of details missing)}$
The coefficient of $x^ky^{n−k}$ is equal to the coefficient of $x^{n−k}y^k$. So we have $n−5=15n−5=15$ and hence $ n=20$.
$underline{text{Other solutions to}:: displaystylebinom{n}{5}=binom{n}{15}quad(**)}$
The binomial coefficient is defined as $$displaystyle binom{n}{k}=begin{cases}dfrac{n!}{k!(n-k)!} &0leq k leq n \ 0 &text{otherwise} end{cases}$$ So by definition, $n=0,1,2,3,4$ are also solutions to $(**)$.
$$displaystyle binom{n}{5}=binom{n}{15}implies displaystyle dfrac{15!}{5!}=dfrac{(n-5)!}{(n-15)!}$$
If $zinmathbb{C}$ and $a,binmathbb{Z}:$ we have $$dfrac{Gamma(z-a+1)}{ Gamma(z-b+1)}=(-1)^{b-a}frac{Gamma(b-z)}{ Gamma(a-z)}$$ Here $a=5, b=15$.
$$dfrac{Gamma(z-5+1)}{ Gamma(z-15+1)}=(-1)^{15-5}frac{Gamma(15-z)}{ Gamma(5-z)}=prod_{i=5}^{14}(i-z)$$
So $displaystyle dfrac{15!}{5!}= prod_{i=5}^{14}(i-z)$ has $10$ roots by fundamental theorem of algebra which are complex (requires more details).
$rule{17cm}{0.5pt}$
So for our question here it is reasonable to assume $n$ to be positive, $ngeq k$ and the expansion of $(x+y)^n$ to have $n+1$ terms.
So $fbox{$n$=$20$}$
answered Nov 26 '18 at 9:18
Yadati KiranYadati Kiran
1,749619
edited Nov 26 '18 at 13:58
answered Nov 26 '18 at 9:18
Yadati KiranYadati Kiran
1,749619
answered Nov 26 '18 at 9:18
Yadati KiranYadati Kiran
1,749619
answered Nov 26 '18 at 9:18
Yadati KiranYadati Kiran
1,749619
1,749619
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Yadati, how did you arrive at n=20.? Am a bit behind please. Usually, I have difficulties dealing with factorials of this nature
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– Bravie
Nov 26 '18 at 9:41
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@NoChance: The question does say the coefficients are equal.
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– Yadati Kiran
Nov 26 '18 at 9:45
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@bof: True, but then you need to expand the binomial coefficient to complex plane through Gamma function, and it can get quite complicated I think.
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– Francis
Nov 26 '18 at 9:46
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The coefficients in the binomial expansion are symmetric - the coefficient of $x^ky^{n-k}$ is equal to the coefficient of $x^{n-k}y^k$. So in this case we must have $n-5=15$ and hence $n=20$.
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– gandalf61
Nov 26 '18 at 9:52
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@gandalf61 You are repeating the argument showing that $n=20$ is a solution, but not advancing one iota in the direction of a proof that it is the only one. As a matter of fact, $n=20$ is not the only solution since $n=0$, $1$, $2$, $3$ and $4$ also are, in contradiction to your argument.
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– Did
Nov 26 '18 at 10:00
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show 10 more comments
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Yadati, how did you arrive at n=20.? Am a bit behind please. Usually, I have difficulties dealing with factorials of this nature
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– Bravie
Nov 26 '18 at 9:41
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@NoChance: The question does say the coefficients are equal.
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– Yadati Kiran
Nov 26 '18 at 9:45
1
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@bof: True, but then you need to expand the binomial coefficient to complex plane through Gamma function, and it can get quite complicated I think.
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– Francis
Nov 26 '18 at 9:46
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The coefficients in the binomial expansion are symmetric - the coefficient of $x^ky^{n-k}$ is equal to the coefficient of $x^{n-k}y^k$. So in this case we must have $n-5=15$ and hence $n=20$.
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– gandalf61
Nov 26 '18 at 9:52
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@gandalf61 You are repeating the argument showing that $n=20$ is a solution, but not advancing one iota in the direction of a proof that it is the only one. As a matter of fact, $n=20$ is not the only solution since $n=0$, $1$, $2$, $3$ and $4$ also are, in contradiction to your argument.
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– Did
Nov 26 '18 at 10:00
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Yadati, how did you arrive at n=20.? Am a bit behind please. Usually, I have difficulties dealing with factorials of this nature
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– Bravie
Nov 26 '18 at 9:41
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Yadati, how did you arrive at n=20.? Am a bit behind please. Usually, I have difficulties dealing with factorials of this nature
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– Bravie
Nov 26 '18 at 9:41
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@NoChance: The question does say the coefficients are equal.
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– Yadati Kiran
Nov 26 '18 at 9:45
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@NoChance: The question does say the coefficients are equal.
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– Yadati Kiran
Nov 26 '18 at 9:45
1
1
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@bof: True, but then you need to expand the binomial coefficient to complex plane through Gamma function, and it can get quite complicated I think.
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– Francis
Nov 26 '18 at 9:46
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@bof: True, but then you need to expand the binomial coefficient to complex plane through Gamma function, and it can get quite complicated I think.
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– Francis
Nov 26 '18 at 9:46
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The coefficients in the binomial expansion are symmetric - the coefficient of $x^ky^{n-k}$ is equal to the coefficient of $x^{n-k}y^k$. So in this case we must have $n-5=15$ and hence $n=20$.
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– gandalf61
Nov 26 '18 at 9:52
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The coefficients in the binomial expansion are symmetric - the coefficient of $x^ky^{n-k}$ is equal to the coefficient of $x^{n-k}y^k$. So in this case we must have $n-5=15$ and hence $n=20$.
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– gandalf61
Nov 26 '18 at 9:52
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@gandalf61 You are repeating the argument showing that $n=20$ is a solution, but not advancing one iota in the direction of a proof that it is the only one. As a matter of fact, $n=20$ is not the only solution since $n=0$, $1$, $2$, $3$ and $4$ also are, in contradiction to your argument.
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– Did
Nov 26 '18 at 10:00
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@gandalf61 You are repeating the argument showing that $n=20$ is a solution, but not advancing one iota in the direction of a proof that it is the only one. As a matter of fact, $n=20$ is not the only solution since $n=0$, $1$, $2$, $3$ and $4$ also are, in contradiction to your argument.
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– Did
Nov 26 '18 at 10:00
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Please edit the question so that it is readable. what is (n 5)? is it nC5? What is (y) 15?
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– NoChance
Nov 26 '18 at 9:21
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i.stack.imgur.com/4jUOr.png
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– Did
Nov 26 '18 at 10:30
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@Did: Sir this is with atmost humility I am asking. If $n=0, n=1, n=2, n=3, n=4$ then what will the $6^{th}$ term and the $16^{th}$ term be if not $ 0$?
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– Yadati Kiran
Nov 26 '18 at 10:44
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@YadatiKiran Again, did I write anything resembling this assertion? Since I have not, why are you asking me this? IOW, what can the phrase
then n=0, n=1, n=2, n=3 and n=4 are solutions, taken from my comment above, possibly mean in your opinion? Sorry but posting random comments will not change the fact that your post below is not a complete answer.$endgroup$
– Did
Nov 26 '18 at 10:46
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Assume one is looking for nonnegative integers, then n=0, n=1, n=2, n=3 and n=4 are solutions. The other solutions solve the identity Γ(a+1)Γ(n−a+1)=Γ(b+1)Γ(n−b+1) for (a,b)=(15,5) It appears that the a−b roots of this are distinct, equal to a+b−12+a+b+12e2iℓπ/(a−b) for ℓ∈{0,1,…,a−b−1}. If this holds, the only real roots are n=a+b and, if a+b is even, n=−1. So the only supplementary nonnegative integer root is n=a+b=20 (Note that none of this seems to be even mentioned [below].) – Did (copy sorry for the lack of format)
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– quid♦
Nov 26 '18 at 12:28