<s>[INST] Find the coefficient of $x^{16}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^4$ We are supposed to find the coefficients of it, I wanted to know if my approach is right here. The final answers seems a bit iffy.
$$(1+x+\dots+x^5)^4=\left(\frac{1-x^6}{1-x}\right)^4=(1-x^6)^4(1-x)^{-4}$$
Using the binomial theoerem, I got the following:
$$
(1-x^6)^4=\sum_{k\ge0}(-1)^k\binom{4}{k}x^{6k}
$$
and using the negative binomial theorem,
$$
(1-x)^{-4}=\sum_{k\ge0}(-1)^k\binom{-4}{k}x^k=\sum_{k\ge0}\binom{4+k-1}{k}x^k
$$
So the $x^{16}$ coefficient is
$$
\binom{4}{0}\binom{8+16-1}{16}-\binom{4}{1}\binom{8+10-1}{10}+\binom{4}{2}\binom{8+4-1}{4}\
$$
$$
 = 1(245157)-4(19448)+6(330)\
$$
$$
= 245157-77792+1980\
= 169345
$$
Not too sure if I did this right. I think my math may be off somewhere but can't figure out where.
 [/INST] $$1+x+x^2+\cdots+x^5=\frac{1-x^{6}}{1-x}=(1-x^{6})(1-x)^{-1}$$
so
$$(1+x+x^2+\cdots+x^5)^4=\left(\frac{1-x^{6}}{1-x}\right)^4=(1-x^{6})^4(1-x)^{-4}=(1-4x^{6}+\cdots)(1+4x+10x^2+20x^3+35x^4+56x^5)$$
There is only one $x^4$ term, and you can see immediately that its coefficient is $35$.
</s>