What is the coefficient of $x^{11}$ in $(3x-9)^{19}$?
$begingroup$
I am currently studying for finals, and I do not know how to do this problem from my study guide. I have tried to watch a few YouTube videos and I know that I will end up with $3x^{11} times (-9)^8$, but from here I do not know where to go. The YouTube videos I have watched have been unclear, and I do not have a chance to ask my professor until just a little before the exam. If anyone has any tips on where to go from here I would greatly appreciate it!
exponential-function binomial-coefficients binomial-theorem
edited Dec 7 '18 at 17:10
PM.
3,3532925
asked Dec 7 '18 at 16:36
bmurf17bmurf17
104
$endgroup$
add a comment |
$begingroup$
I am currently studying for finals, and I do not know how to do this problem from my study guide. I have tried to watch a few YouTube videos and I know that I will end up with $3x^{11} times (-9)^8$, but from here I do not know where to go. The YouTube videos I have watched have been unclear, and I do not have a chance to ask my professor until just a little before the exam. If anyone has any tips on where to go from here I would greatly appreciate it!
exponential-function binomial-coefficients binomial-theorem
edited Dec 7 '18 at 17:10
PM.
3,3532925
asked Dec 7 '18 at 16:36
bmurf17bmurf17
104
$endgroup$
$begingroup$
Don't forget about the binomial coefficient.
$endgroup$
– John Wayland Bales
Dec 7 '18 at 16:40
$begingroup$
Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
$endgroup$
– Prakhar Nagpal
Dec 7 '18 at 16:41
add a comment |
$begingroup$
I am currently studying for finals, and I do not know how to do this problem from my study guide. I have tried to watch a few YouTube videos and I know that I will end up with $3x^{11} times (-9)^8$, but from here I do not know where to go. The YouTube videos I have watched have been unclear, and I do not have a chance to ask my professor until just a little before the exam. If anyone has any tips on where to go from here I would greatly appreciate it!
exponential-function binomial-coefficients binomial-theorem
edited Dec 7 '18 at 17:10
PM.
3,3532925
asked Dec 7 '18 at 16:36
bmurf17bmurf17
104
$endgroup$
I am currently studying for finals, and I do not know how to do this problem from my study guide. I have tried to watch a few YouTube videos and I know that I will end up with $3x^{11} times (-9)^8$, but from here I do not know where to go. The YouTube videos I have watched have been unclear, and I do not have a chance to ask my professor until just a little before the exam. If anyone has any tips on where to go from here I would greatly appreciate it!
exponential-function binomial-coefficients binomial-theorem
exponential-function binomial-coefficients binomial-theorem
edited Dec 7 '18 at 17:10
PM.
3,3532925
asked Dec 7 '18 at 16:36
bmurf17bmurf17
104
edited Dec 7 '18 at 17:10
PM.
3,3532925
asked Dec 7 '18 at 16:36
bmurf17bmurf17
104
edited Dec 7 '18 at 17:10
PM.
3,3532925
edited Dec 7 '18 at 17:10
PM.
3,3532925
edited Dec 7 '18 at 17:10
PM.
3,3532925
3,3532925
asked Dec 7 '18 at 16:36
bmurf17bmurf17
104
asked Dec 7 '18 at 16:36
bmurf17bmurf17
104
asked Dec 7 '18 at 16:36
bmurf17bmurf17
104
104
$begingroup$
Don't forget about the binomial coefficient.
$endgroup$
– John Wayland Bales
Dec 7 '18 at 16:40
$begingroup$
Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
$endgroup$
– Prakhar Nagpal
Dec 7 '18 at 16:41
add a comment |
$begingroup$
Don't forget about the binomial coefficient.
$endgroup$
– John Wayland Bales
Dec 7 '18 at 16:40
$begingroup$
Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
$endgroup$
– Prakhar Nagpal
Dec 7 '18 at 16:41
$begingroup$
Don't forget about the binomial coefficient.
$endgroup$
– John Wayland Bales
Dec 7 '18 at 16:40
$begingroup$
Don't forget about the binomial coefficient.
$endgroup$
– John Wayland Bales
Dec 7 '18 at 16:40
$begingroup$
Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
$endgroup$
– Prakhar Nagpal
Dec 7 '18 at 16:41
$begingroup$
Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
$endgroup$
– Prakhar Nagpal
Dec 7 '18 at 16:41
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$$(3x-9)^{19} = (3x-9)(3x-9) cdots (3x-9) $$
Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.
If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.
If you choose $3x$ for $11$ times, you end up with $color{blue}{(3x)^{11}}$, and you can do this in exactly $color{green}{binom{19}{11}}$ ways. $color{grey}{text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$
For each of the above $color{green}{binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $color{red}{(-9)^8}$
So the overall term would be $color{green}{binom{19}{11}}color{blue}{(3x)^{11}}color{red}{(-9)^8}$
answered Dec 7 '18 at 17:00
rsadhvikarsadhvika
1,7101228
$endgroup$
$begingroup$
Thank you this makes sense! I completely forgot about the choose
$endgroup$
– bmurf17
Dec 7 '18 at 19:04
add a comment |
$begingroup$
HINT:
The Binomial expansion : $$(a+b)^{n}= sum_{k=0}^{n} {n choose k}{a}^{k}{b}^{n-k}.$$
Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n choose k}{a}^{k}{b}^{n-k}$
answered Dec 7 '18 at 16:48
Thomas ShelbyThomas Shelby
3,9542625
$endgroup$
add a comment |
$begingroup$
Note that by expansion of a binomial to the $n^{th}$ power, you get
$$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$
where ${n choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)
For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get
$${19 choose 19-8} (3x)^{11}(-9)^8$$
$${19 choose 11} (3x)^{11}9^8$$
Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.
answered Dec 7 '18 at 16:44
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
You are part way there.
If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.
There are $19 choose 11$ ways of doing that, so the term in $x^{11}$ is
$$
{19 choose 11} (3x)^{11} (-9)^8
$$
Thus the coefficient of $x^{11}$ is
$$
{19 choose 11} (3)^{11} (-9)^8 = {19 choose 11}3^{27}
$$
answered Dec 7 '18 at 17:04
PM.PM.
3,3532925
$endgroup$
add a comment |
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4 Answers
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4 Answers
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$begingroup$
$$(3x-9)^{19} = (3x-9)(3x-9) cdots (3x-9) $$
Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.
If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.
If you choose $3x$ for $11$ times, you end up with $color{blue}{(3x)^{11}}$, and you can do this in exactly $color{green}{binom{19}{11}}$ ways. $color{grey}{text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$
For each of the above $color{green}{binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $color{red}{(-9)^8}$
So the overall term would be $color{green}{binom{19}{11}}color{blue}{(3x)^{11}}color{red}{(-9)^8}$
answered Dec 7 '18 at 17:00
rsadhvikarsadhvika
1,7101228
$endgroup$
$begingroup$
Thank you this makes sense! I completely forgot about the choose
$endgroup$
– bmurf17
Dec 7 '18 at 19:04
add a comment |
$begingroup$
$$(3x-9)^{19} = (3x-9)(3x-9) cdots (3x-9) $$
Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.
If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.
If you choose $3x$ for $11$ times, you end up with $color{blue}{(3x)^{11}}$, and you can do this in exactly $color{green}{binom{19}{11}}$ ways. $color{grey}{text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$
For each of the above $color{green}{binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $color{red}{(-9)^8}$
So the overall term would be $color{green}{binom{19}{11}}color{blue}{(3x)^{11}}color{red}{(-9)^8}$
answered Dec 7 '18 at 17:00
rsadhvikarsadhvika
1,7101228
$endgroup$
$begingroup$
Thank you this makes sense! I completely forgot about the choose
$endgroup$
– bmurf17
Dec 7 '18 at 19:04
add a comment |
$begingroup$
$$(3x-9)^{19} = (3x-9)(3x-9) cdots (3x-9) $$
Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.
If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.
If you choose $3x$ for $11$ times, you end up with $color{blue}{(3x)^{11}}$, and you can do this in exactly $color{green}{binom{19}{11}}$ ways. $color{grey}{text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$
For each of the above $color{green}{binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $color{red}{(-9)^8}$
So the overall term would be $color{green}{binom{19}{11}}color{blue}{(3x)^{11}}color{red}{(-9)^8}$
answered Dec 7 '18 at 17:00
rsadhvikarsadhvika
1,7101228
$endgroup$
$$(3x-9)^{19} = (3x-9)(3x-9) cdots (3x-9) $$
Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.
If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.
If you choose $3x$ for $11$ times, you end up with $color{blue}{(3x)^{11}}$, and you can do this in exactly $color{green}{binom{19}{11}}$ ways. $color{grey}{text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$
For each of the above $color{green}{binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $color{red}{(-9)^8}$
So the overall term would be $color{green}{binom{19}{11}}color{blue}{(3x)^{11}}color{red}{(-9)^8}$
answered Dec 7 '18 at 17:00
rsadhvikarsadhvika
1,7101228
edited Dec 7 '18 at 17:12
answered Dec 7 '18 at 17:00
rsadhvikarsadhvika
1,7101228
answered Dec 7 '18 at 17:00
rsadhvikarsadhvika
1,7101228
answered Dec 7 '18 at 17:00
rsadhvikarsadhvika
1,7101228
1,7101228
$begingroup$
Thank you this makes sense! I completely forgot about the choose
$endgroup$
– bmurf17
Dec 7 '18 at 19:04
add a comment |
$begingroup$
Thank you this makes sense! I completely forgot about the choose
$endgroup$
– bmurf17
Dec 7 '18 at 19:04
$begingroup$
Thank you this makes sense! I completely forgot about the choose
$endgroup$
– bmurf17
Dec 7 '18 at 19:04
$begingroup$
Thank you this makes sense! I completely forgot about the choose
$endgroup$
– bmurf17
Dec 7 '18 at 19:04
add a comment |
$begingroup$
HINT:
The Binomial expansion : $$(a+b)^{n}= sum_{k=0}^{n} {n choose k}{a}^{k}{b}^{n-k}.$$
Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n choose k}{a}^{k}{b}^{n-k}$
answered Dec 7 '18 at 16:48
Thomas ShelbyThomas Shelby
3,9542625
$endgroup$
add a comment |
$begingroup$
HINT:
The Binomial expansion : $$(a+b)^{n}= sum_{k=0}^{n} {n choose k}{a}^{k}{b}^{n-k}.$$
Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n choose k}{a}^{k}{b}^{n-k}$
answered Dec 7 '18 at 16:48
Thomas ShelbyThomas Shelby
3,9542625
$endgroup$
add a comment |
$begingroup$
HINT:
The Binomial expansion : $$(a+b)^{n}= sum_{k=0}^{n} {n choose k}{a}^{k}{b}^{n-k}.$$
Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n choose k}{a}^{k}{b}^{n-k}$
answered Dec 7 '18 at 16:48
Thomas ShelbyThomas Shelby
3,9542625
$endgroup$
HINT:
The Binomial expansion : $$(a+b)^{n}= sum_{k=0}^{n} {n choose k}{a}^{k}{b}^{n-k}.$$
Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n choose k}{a}^{k}{b}^{n-k}$
answered Dec 7 '18 at 16:48
Thomas ShelbyThomas Shelby
3,9542625
answered Dec 7 '18 at 16:48
Thomas ShelbyThomas Shelby
3,9542625
answered Dec 7 '18 at 16:48
Thomas ShelbyThomas Shelby
3,9542625
answered Dec 7 '18 at 16:48
Thomas ShelbyThomas Shelby
3,9542625
3,9542625
add a comment |
add a comment |
$begingroup$
Note that by expansion of a binomial to the $n^{th}$ power, you get
$$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$
where ${n choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)
For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get
$${19 choose 19-8} (3x)^{11}(-9)^8$$
$${19 choose 11} (3x)^{11}9^8$$
Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.
answered Dec 7 '18 at 16:44
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
Note that by expansion of a binomial to the $n^{th}$ power, you get
$$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$
where ${n choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)
For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get
$${19 choose 19-8} (3x)^{11}(-9)^8$$
$${19 choose 11} (3x)^{11}9^8$$
Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.
answered Dec 7 '18 at 16:44
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
Note that by expansion of a binomial to the $n^{th}$ power, you get
$$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$
where ${n choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)
For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get
$${19 choose 19-8} (3x)^{11}(-9)^8$$
$${19 choose 11} (3x)^{11}9^8$$
Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.
answered Dec 7 '18 at 16:44
KM101KM101
6,0901525
$endgroup$
Note that by expansion of a binomial to the $n^{th}$ power, you get
$$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$
where ${n choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)
For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get
$${19 choose 19-8} (3x)^{11}(-9)^8$$
$${19 choose 11} (3x)^{11}9^8$$
Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.
answered Dec 7 '18 at 16:44
KM101KM101
6,0901525
edited Dec 7 '18 at 17:02
answered Dec 7 '18 at 16:44
KM101KM101
6,0901525
answered Dec 7 '18 at 16:44
KM101KM101
6,0901525
answered Dec 7 '18 at 16:44
KM101KM101
6,0901525
6,0901525
add a comment |
add a comment |
$begingroup$
You are part way there.
If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.
There are $19 choose 11$ ways of doing that, so the term in $x^{11}$ is
$$
{19 choose 11} (3x)^{11} (-9)^8
$$
Thus the coefficient of $x^{11}$ is
$$
{19 choose 11} (3)^{11} (-9)^8 = {19 choose 11}3^{27}
$$
answered Dec 7 '18 at 17:04
PM.PM.
3,3532925
$endgroup$
add a comment |
$begingroup$
You are part way there.
If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.
There are $19 choose 11$ ways of doing that, so the term in $x^{11}$ is
$$
{19 choose 11} (3x)^{11} (-9)^8
$$
Thus the coefficient of $x^{11}$ is
$$
{19 choose 11} (3)^{11} (-9)^8 = {19 choose 11}3^{27}
$$
answered Dec 7 '18 at 17:04
PM.PM.
3,3532925
$endgroup$
add a comment |
$begingroup$
You are part way there.
If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.
There are $19 choose 11$ ways of doing that, so the term in $x^{11}$ is
$$
{19 choose 11} (3x)^{11} (-9)^8
$$
Thus the coefficient of $x^{11}$ is
$$
{19 choose 11} (3)^{11} (-9)^8 = {19 choose 11}3^{27}
$$
answered Dec 7 '18 at 17:04
PM.PM.
3,3532925
$endgroup$
You are part way there.
If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.
There are $19 choose 11$ ways of doing that, so the term in $x^{11}$ is
$$
{19 choose 11} (3x)^{11} (-9)^8
$$
Thus the coefficient of $x^{11}$ is
$$
{19 choose 11} (3)^{11} (-9)^8 = {19 choose 11}3^{27}
$$
answered Dec 7 '18 at 17:04
PM.PM.
3,3532925
answered Dec 7 '18 at 17:04
PM.PM.
3,3532925
answered Dec 7 '18 at 17:04
PM.PM.
3,3532925
answered Dec 7 '18 at 17:04
PM.PM.
3,3532925
3,3532925
add a comment |
add a comment |
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$begingroup$
Don't forget about the binomial coefficient.
$endgroup$
– John Wayland Bales
Dec 7 '18 at 16:40
$begingroup$
Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
$endgroup$
– Prakhar Nagpal
Dec 7 '18 at 16:41