What is the coefficient of $x^{100}$ when expanding $(1+x+x^2+x^3+dots+x^{100})^3$?
up vote
2
down vote
favorite
What is the coefficient of $x^{100}$ when expanding $(1+x+x^2+x^3+dots+x^{100})^3$?
I tried on this question and have attempted many different methods. Every time I get a different answer and my solution is usually long. I hope there is a simple method for answering this question, thank you.
algebra-precalculus
edited 21 hours ago
Robert Z
91.1k1058129
asked 22 hours ago
user587054
1749
add a comment |
up vote
2
down vote
favorite
What is the coefficient of $x^{100}$ when expanding $(1+x+x^2+x^3+dots+x^{100})^3$?
I tried on this question and have attempted many different methods. Every time I get a different answer and my solution is usually long. I hope there is a simple method for answering this question, thank you.
algebra-precalculus
edited 21 hours ago
Robert Z
91.1k1058129
asked 22 hours ago
user587054
1749
Do you remember how, in expanding $(a+b)^n$, we use the binomial coefficient $binom{n}{k}$ to determine the $k$-th coefficient? There's an analogoue for various other things in the parentheses as a sort of generalization - multinomials, the multinomial theorem, and the multinomial coefficient. en.wikipedia.org/wiki/Multinomial_theorem This might worth reading up on - I don't know enough to be able to properly help you with this specific question aside that I know this thing exists, and I believe lab bhattacharjee already provided a sufficient answer.
– Eevee Trainer
22 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
What is the coefficient of $x^{100}$ when expanding $(1+x+x^2+x^3+dots+x^{100})^3$?
I tried on this question and have attempted many different methods. Every time I get a different answer and my solution is usually long. I hope there is a simple method for answering this question, thank you.
algebra-precalculus
edited 21 hours ago
Robert Z
91.1k1058129
asked 22 hours ago
user587054
1749
What is the coefficient of $x^{100}$ when expanding $(1+x+x^2+x^3+dots+x^{100})^3$?
I tried on this question and have attempted many different methods. Every time I get a different answer and my solution is usually long. I hope there is a simple method for answering this question, thank you.
algebra-precalculus
algebra-precalculus
edited 21 hours ago
Robert Z
91.1k1058129
asked 22 hours ago
user587054
1749
edited 21 hours ago
Robert Z
91.1k1058129
asked 22 hours ago
user587054
1749
edited 21 hours ago
Robert Z
91.1k1058129
edited 21 hours ago
Robert Z
91.1k1058129
edited 21 hours ago
Robert Z
91.1k1058129
91.1k1058129
asked 22 hours ago
user587054
1749
asked 22 hours ago
user587054
1749
asked 22 hours ago
user587054
1749
1749
Do you remember how, in expanding $(a+b)^n$, we use the binomial coefficient $binom{n}{k}$ to determine the $k$-th coefficient? There's an analogoue for various other things in the parentheses as a sort of generalization - multinomials, the multinomial theorem, and the multinomial coefficient. en.wikipedia.org/wiki/Multinomial_theorem This might worth reading up on - I don't know enough to be able to properly help you with this specific question aside that I know this thing exists, and I believe lab bhattacharjee already provided a sufficient answer.
– Eevee Trainer
22 hours ago
add a comment |
Do you remember how, in expanding $(a+b)^n$, we use the binomial coefficient $binom{n}{k}$ to determine the $k$-th coefficient? There's an analogoue for various other things in the parentheses as a sort of generalization - multinomials, the multinomial theorem, and the multinomial coefficient. en.wikipedia.org/wiki/Multinomial_theorem This might worth reading up on - I don't know enough to be able to properly help you with this specific question aside that I know this thing exists, and I believe lab bhattacharjee already provided a sufficient answer.
– Eevee Trainer
22 hours ago
Do you remember how, in expanding $(a+b)^n$, we use the binomial coefficient $binom{n}{k}$ to determine the $k$-th coefficient? There's an analogoue for various other things in the parentheses as a sort of generalization - multinomials, the multinomial theorem, and the multinomial coefficient. en.wikipedia.org/wiki/Multinomial_theorem This might worth reading up on - I don't know enough to be able to properly help you with this specific question aside that I know this thing exists, and I believe lab bhattacharjee already provided a sufficient answer.
– Eevee Trainer
22 hours ago
Do you remember how, in expanding $(a+b)^n$, we use the binomial coefficient $binom{n}{k}$ to determine the $k$-th coefficient? There's an analogoue for various other things in the parentheses as a sort of generalization - multinomials, the multinomial theorem, and the multinomial coefficient. en.wikipedia.org/wiki/Multinomial_theorem This might worth reading up on - I don't know enough to be able to properly help you with this specific question aside that I know this thing exists, and I believe lab bhattacharjee already provided a sufficient answer.
– Eevee Trainer
22 hours ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
A direct counting argument works as well. Note that the $x^{100}$ term of the product will be the number of ways to get $x^{100} = x^ax^bx^c$, where $0 leq a,b,c leq 100$ (here, the $x^a$ represents the contribution of the first $(1 + x + dots + x^{100})$, $x^b$ the contribution of the second term, and $x^c$ for the third).
But this is just asking "how many ways are there to choose $a,b,c$ all nonnegative such that $a + b + c = 100$?" This can be solved by direct application of stars and bars to get $binom{102}{2} = boxed{5151}$.
edited 21 hours ago
Robert Z
91.1k1058129
answered 22 hours ago
platty
2,323215
2
Nice and clear answer (+1)
– Robert Z
21 hours ago
add a comment |
up vote
9
down vote
Hint:
For $xne1,$
$$(1+x+x^2+cdots+x^{100})^3=(1-x^{101})^3(1-x)^{-3}$$
So, we require the coefficient of $x^{100}$ in $1cdot(1-x)^{-3}$
Now use Binomial series
answered 22 hours ago
lab bhattacharjee
221k15154271
1
Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
– lab bhattacharjee
22 hours ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
A direct counting argument works as well. Note that the $x^{100}$ term of the product will be the number of ways to get $x^{100} = x^ax^bx^c$, where $0 leq a,b,c leq 100$ (here, the $x^a$ represents the contribution of the first $(1 + x + dots + x^{100})$, $x^b$ the contribution of the second term, and $x^c$ for the third).
But this is just asking "how many ways are there to choose $a,b,c$ all nonnegative such that $a + b + c = 100$?" This can be solved by direct application of stars and bars to get $binom{102}{2} = boxed{5151}$.
edited 21 hours ago
Robert Z
91.1k1058129
answered 22 hours ago
platty
2,323215
2
Nice and clear answer (+1)
– Robert Z
21 hours ago
add a comment |
up vote
8
down vote
accepted
A direct counting argument works as well. Note that the $x^{100}$ term of the product will be the number of ways to get $x^{100} = x^ax^bx^c$, where $0 leq a,b,c leq 100$ (here, the $x^a$ represents the contribution of the first $(1 + x + dots + x^{100})$, $x^b$ the contribution of the second term, and $x^c$ for the third).
But this is just asking "how many ways are there to choose $a,b,c$ all nonnegative such that $a + b + c = 100$?" This can be solved by direct application of stars and bars to get $binom{102}{2} = boxed{5151}$.
edited 21 hours ago
Robert Z
91.1k1058129
answered 22 hours ago
platty
2,323215
2
Nice and clear answer (+1)
– Robert Z
21 hours ago
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
A direct counting argument works as well. Note that the $x^{100}$ term of the product will be the number of ways to get $x^{100} = x^ax^bx^c$, where $0 leq a,b,c leq 100$ (here, the $x^a$ represents the contribution of the first $(1 + x + dots + x^{100})$, $x^b$ the contribution of the second term, and $x^c$ for the third).
But this is just asking "how many ways are there to choose $a,b,c$ all nonnegative such that $a + b + c = 100$?" This can be solved by direct application of stars and bars to get $binom{102}{2} = boxed{5151}$.
edited 21 hours ago
Robert Z
91.1k1058129
answered 22 hours ago
platty
2,323215
A direct counting argument works as well. Note that the $x^{100}$ term of the product will be the number of ways to get $x^{100} = x^ax^bx^c$, where $0 leq a,b,c leq 100$ (here, the $x^a$ represents the contribution of the first $(1 + x + dots + x^{100})$, $x^b$ the contribution of the second term, and $x^c$ for the third).
But this is just asking "how many ways are there to choose $a,b,c$ all nonnegative such that $a + b + c = 100$?" This can be solved by direct application of stars and bars to get $binom{102}{2} = boxed{5151}$.
edited 21 hours ago
Robert Z
91.1k1058129
answered 22 hours ago
platty
2,323215
edited 21 hours ago
Robert Z
91.1k1058129
edited 21 hours ago
Robert Z
91.1k1058129
edited 21 hours ago
Robert Z
91.1k1058129
91.1k1058129
answered 22 hours ago
platty
2,323215
answered 22 hours ago
platty
2,323215
answered 22 hours ago
platty
2,323215
2,323215
2
Nice and clear answer (+1)
– Robert Z
21 hours ago
add a comment |
2
Nice and clear answer (+1)
– Robert Z
21 hours ago
2
2
Nice and clear answer (+1)
– Robert Z
21 hours ago
Nice and clear answer (+1)
– Robert Z
21 hours ago
add a comment |
up vote
9
down vote
Hint:
For $xne1,$
$$(1+x+x^2+cdots+x^{100})^3=(1-x^{101})^3(1-x)^{-3}$$
So, we require the coefficient of $x^{100}$ in $1cdot(1-x)^{-3}$
Now use Binomial series
answered 22 hours ago
lab bhattacharjee
221k15154271
1
Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
– lab bhattacharjee
22 hours ago
add a comment |
up vote
9
down vote
Hint:
For $xne1,$
$$(1+x+x^2+cdots+x^{100})^3=(1-x^{101})^3(1-x)^{-3}$$
So, we require the coefficient of $x^{100}$ in $1cdot(1-x)^{-3}$
Now use Binomial series
answered 22 hours ago
lab bhattacharjee
221k15154271
1
Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
– lab bhattacharjee
22 hours ago
add a comment |
up vote
9
down vote
up vote
9
down vote
Hint:
For $xne1,$
$$(1+x+x^2+cdots+x^{100})^3=(1-x^{101})^3(1-x)^{-3}$$
So, we require the coefficient of $x^{100}$ in $1cdot(1-x)^{-3}$
Now use Binomial series
answered 22 hours ago
lab bhattacharjee
221k15154271
Hint:
For $xne1,$
$$(1+x+x^2+cdots+x^{100})^3=(1-x^{101})^3(1-x)^{-3}$$
So, we require the coefficient of $x^{100}$ in $1cdot(1-x)^{-3}$
Now use Binomial series
answered 22 hours ago
lab bhattacharjee
221k15154271
answered 22 hours ago
lab bhattacharjee
221k15154271
answered 22 hours ago
lab bhattacharjee
221k15154271
answered 22 hours ago
lab bhattacharjee
221k15154271
221k15154271
1
Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
– lab bhattacharjee
22 hours ago
add a comment |
1
Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
– lab bhattacharjee
22 hours ago
1
1
Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
– lab bhattacharjee
22 hours ago
Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
– lab bhattacharjee
22 hours ago
add a comment |
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Do you remember how, in expanding $(a+b)^n$, we use the binomial coefficient $binom{n}{k}$ to determine the $k$-th coefficient? There's an analogoue for various other things in the parentheses as a sort of generalization - multinomials, the multinomial theorem, and the multinomial coefficient. en.wikipedia.org/wiki/Multinomial_theorem This might worth reading up on - I don't know enough to be able to properly help you with this specific question aside that I know this thing exists, and I believe lab bhattacharjee already provided a sufficient answer.
– Eevee Trainer
22 hours ago