What is the coefficient of $x^{100}$ when expanding $(1+x+x^2+x^3+dots+x^{100})^3$?












2












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Nov 30 '18 at 7:21
















2












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Nov 30 '18 at 7:21














2












2








2





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 7:41









Robert Z

94.2k1061132




94.2k1061132










asked Nov 30 '18 at 7:13









user587054user587054

46011




46011












  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Nov 30 '18 at 7:21


















  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Nov 30 '18 at 7:21
















$begingroup$
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.
$endgroup$
– Eevee Trainer
Nov 30 '18 at 7:21




$begingroup$
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.
$endgroup$
– Eevee Trainer
Nov 30 '18 at 7:21










2 Answers
2






active

oldest

votes


















8












$begingroup$

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 $binom{102}{2} = boxed{5151}$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Nice and clear answer (+1)
    $endgroup$
    – Robert Z
    Nov 30 '18 at 7:37



















9












$begingroup$

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






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 7:29











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

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 $binom{102}{2} = boxed{5151}$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Nice and clear answer (+1)
    $endgroup$
    – Robert Z
    Nov 30 '18 at 7:37
















8












$begingroup$

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 $binom{102}{2} = boxed{5151}$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Nice and clear answer (+1)
    $endgroup$
    – Robert Z
    Nov 30 '18 at 7:37














8












8








8





$begingroup$

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 $binom{102}{2} = boxed{5151}$.






share|cite|improve this answer











$endgroup$



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 $binom{102}{2} = boxed{5151}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 7:39









Robert Z

94.2k1061132




94.2k1061132










answered Nov 30 '18 at 7:27









plattyplatty

3,370320




3,370320








  • 2




    $begingroup$
    Nice and clear answer (+1)
    $endgroup$
    – Robert Z
    Nov 30 '18 at 7:37














  • 2




    $begingroup$
    Nice and clear answer (+1)
    $endgroup$
    – Robert Z
    Nov 30 '18 at 7:37








2




2




$begingroup$
Nice and clear answer (+1)
$endgroup$
– Robert Z
Nov 30 '18 at 7:37




$begingroup$
Nice and clear answer (+1)
$endgroup$
– Robert Z
Nov 30 '18 at 7:37











9












$begingroup$

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






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 7:29
















9












$begingroup$

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






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 7:29














9












9








9





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 '18 at 7:18









lab bhattacharjeelab bhattacharjee

224k15156274




224k15156274








  • 1




    $begingroup$
    Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 7:29














  • 1




    $begingroup$
    Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 7:29








1




1




$begingroup$
Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 7:29




$begingroup$
Interestingly the general term of $$(1-x)^{-3}$$ is $$dfrac{(-3)(-4)cdots(-r-1)(-r-2)}{r!}(-x)^r=dfrac{(r+1)(r+2)}2x^r$$
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 7:29


















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