<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] $$y=\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$
the first derivative is
$$\frac{-1}{(1-x)^2}=\sum_{n=0}^{\infty }nx^{n-1}$$
the fourth derivative 
$$\frac{24}{(1-x)^5}=\sum_{n=4}^{\infty }n(n-1)(n-2)(n-3)x^{n-4}$$
let $x\rightarrow x^2$
$$\frac{24}{(1-x^2)^5}=\sum_{n=4}^{\infty }n(n-1)(n-2)(n-3)x^{2n-8}$$
to find coefficient of $x^{12}$
let
$$2n-8=12$$
$$n=10$$
so the coefficient is
$$(10)(10-1)(10-2)(10-3)/24=210$$ 
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