<s>[INST] Find the coefficient of $ x^{12}$ in $(1-x^2)^{-5}$ Find the coefficient of $x^{12}$ in $(1-x^2)^{-5}$
What can be said for $x^{17}$
Tried $\frac{1}{(1-x^2)^{5}}$=$\sum_{n=0}^\infty \binom{n+5-1}{n}x^n$
not sure that i can do that with $x^2$ 
 [/INST] It's convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
[x^{12}]\frac{1}{(1-x^2)^5}
&=[x^{12}]\sum_{k=0}^\infty \binom{-5}{k}(-x^2)^{k}\tag{1}\\
&=[x^{12}]\sum_{k=0}^\infty \binom{k+4}{4}x^{2k}\tag{2}\\
&=\binom{10}{4}\tag{3}\\
&=210
\end{align*}

Comment:


*

*In (1) we use the binomial series expansion of $\frac{1}{(1-x^2)^5}$

*In (2) we use the binomial identity
\begin{align*}
\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k
\end{align*}

*In (3) we select the coefficient of the series with $k=6$ accordingly


Note: Since the function is even, the coefficient of $x^{17}$ is zero.
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