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**Prove It** For starters, you should know that a MacLaurin series for a function f(x) looks like $\displaystyle \displaystyle \begin{align*} f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(0)}{n!}\, x^n \end{align*}$.

Now if $\displaystyle \displaystyle f(x) = \frac{x^2}{1 - x^2} = x^2 \left( \frac{1}{1 - x^2} \right) $ we can recognise that $\displaystyle \displaystyle \frac{1}{1 - x^2}$ is the closed form of the geometric series $\displaystyle \sum_{m = 0}^{\infty} \left( x^2 \right) ^m $ with $\displaystyle \displaystyle \left| x^2 \right| < 1 \implies |x| < 1 $, so that means

$\displaystyle \displaystyle \begin{align*} f(x) &= \frac{x^2}{1 - x^2} \\ &= x^2 \sum_{m = 0}^{\infty} \left( x^2 \right) ^m \\ &= \sum_{m = 0}^{\infty} x^{2m + 2} \\ &= 0 + 0x + 1x^2 + 0x^3 + 1x^4 + 0x^5 + 1x^6 + \dots \end{align*}$

which is a MacLaurin Series. It can be seen that the coefficient of every even-powered term (except the constant term) is 1, so that means the coefficient of the 2012th term is 1. Remember that every coefficient in the MacLaurin Series is of the form $\displaystyle \displaystyle \frac{f^{(n)}(0)}{n!}$, this means for the 2012th term we have

$\displaystyle \displaystyle \begin{align*} \frac{f^{(2012)}(0)}{2012!} &= 1 \\ f^{(2012)}(0) &= 2012! \end{align*}$