Coefficient of $x^{12}$ in $(1+x^2+x^4+x^6)^n$ I need to find the coefficient of $x^{12}$ in the polynomial $(1+x^2+x^4+x^6)^n$.
I have reduced the polynomial to $\left(\frac{1-x^8}{1-x^2}\right)^ n$ and tried binomial expansion and Taylor series, yet it seems too complicated to be worked out by hand.
What should I do?

Your approach is also fine. In the following it is convenient to denote with $[x^k]$ the coefficient of $x^k$ in a series.

We obtain
\begin{align*}
\color{blue}{[x^{12}]}&\color{blue}{\left(\frac{1-x^8}{1-x^2}\right)^n}\\
&=[x^{12}](1-x^8)^n\sum_{j=0}^{\infty}\binom{-n}{j}\left(-x^2\right)^j\tag{1}\\
&=[x^{12}]\left(1-\binom{n}{1}x^8\right)\sum_{j=0}^{\infty}\binom{n+j-1}{j}x^{2j}\tag{2}\\
&=\left([x^{12}]-n[x^4]\right)\sum_{j=0}^{\infty}\binom{n+j-1}{j}x^{2j}\tag{3}\\
&\,\,\color{blue}{=\binom{n+5}{6}-n\binom{n+1}{2}}\tag{4}
\end{align*}

Comment:

*

*In (1) we use the binomial series expansion.


*In (2) we expand $(1-x^8)^n$ up to terms of $x^8$, since other terms do not contribute. We also use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.


*In (3) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.


*In (4) we select the coefficients of $x^k$ accordingly.
