Find the coefficient of $x^{24}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^8$

Question

I'm not sure how to go about doing this. Do I find the ways to add up to 24 using the exponents with repetition? Is the multinomial theorem useful here? I also have a feeling that generating functions might be useful here, but I can't see how. Any help would be appreciated.

Answer 1

$$(1+x+\dots+x^5)^8=\left(\frac{1-x^6}{1-x}\right)^8=(1-x^6)^8(1-x)^{-8}$$ Using the binomial theoerem, $$ (1-x^6)^8=\sum_{k\ge0}(-1)^k\binom{8}{k}x^{6k} $$ and using the negative binomial theorem, $$ (1-x)^{-8}=\sum_{k\ge0}(-1)^k\binom{-8}{k}x^k=\sum_{k\ge0}\binom{8+k-1}{k}x^k $$ Thus, when we convolve the above two generating functions, the $x^{24}$ coefficient is $$ \binom{8}{0}\binom{8+24-1}{24}-\binom{8}{1}\binom{8+18-1}{18}+\binom{8}{2}\binom{8+12-1}{12}\\ -\binom{8}{3}\binom{8+6-1}{6}+\binom{8}{4}\binom{8+0-1}{0} $$ Addendum: If $a(x)=\sum_{n\ge0}a_nx^n$ and $b(x)=\sum_{n\ge 0}b_nx^n$, then the $x^{24}$ coefficient of $c(x)=a(x)b(x)$ is $$ \sum_{k=0}^{24}a _kb_{n-k} $$ The final answer I wrote then comes from setting $a(x)=(1-x^6)^8$, $b(x)=(1-x)^{-8}$, and realizing that $a_k=0$ unless $k$ is a multiple of 6, so the above can be rewritten $$ \sum_{\ell=0}^{4}a _{6\ell}b_{n-6\ell} $$

Answer 2

$\bf{My\; Solution::}$ Let $S = 1+x+x^2+x^3+........+x^5......(1)$

Multiply both side by $x\;,$ We get

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;xS = x+x^2+x^3+..............x^6........(2)$

Now Subtract $(1)$ and $(2)\;,$ we get

$\Rightarrow \displaystyle S(1-x) = 1-x^6\Rightarrow S = \frac{(1-x^6)}{(1-x)}$

So we have to find Coeff. of $x^{24}$ in $\displaystyle (1-x^6)^8\cdot (1-x)^8$

Now Using Binomial Theorem, For $(+ve)$ Integral Index

$\displaystyle (1-y)^n = \binom{n}{0}-\binom{n}{1}y+\binom{n}{2}y^2+..............+(-1)^n\binom{n}{n}y^n$

and Using Binomial Theorem, For $(-ve)$ Integral Index

$\displaystyle (1-y)^{-n} = 1+ny+\frac{n(n+1)}{2}y^2+\frac{n(n+1)(n+2)}{3}y^3+.........$