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.

\begin{align} \left(1+x+x^2+x^3+x^4+x^5\right)^2\ \times \left(1+x+x^2+x^3+x^4+x^5\right)^2 \\
= \left(1+2x+3x^2+4x^3+5x^4+6x^5+\color{red}{5x^6}+\color{orange}{4x^7}+\color{green}{3x^8}+\color{blue}{2x^9}+\color{purple}{x^{10}}\right)\ \times \left(1+2x+3x^2+4x^3+5x^4+6x^5+\color{purple}{5x^6}+\color{blue}{4x^7}+\color{green}{3x^8}+\color{orange}{2x^9}+\color{red}{x^{10}}\right)\\
\\
\end{align}
So the coeffecient of $x^{16}$ is $\ \color{red}{5\times 1} + \color{orange}{4\times 2} + \color{green}{3\times 3} + \color{blue}{2\times 4} + \color{purple}{1\times 5} = 35.$
