Find the power series representation of $\frac{3x^2}{(1-x)^2}$. I'm trying to find the power series representation for the function $\frac{3x^2}{(1-x)^2}$. So far my work has gotten me to this point:
$$\frac{3x^2}{(1-x)^2}={\left(3x\left(\frac{1}{1-x}\right)\right)}^2={\left(3x\sum_{n=0}^{\infty}{x^n}\right)}^2$$
Is this math correct so far? Where do I take it from here if it is?
 A: That's not really the approach you want, and you do have a mistake, because $3^2\ne 3$. It's also (in many cases) a real pain to square a power series. Better to do it this way:
$\begin{align}
f(x)=\frac{3x^2}{(1-x)^2}=3x^2\cdot\frac{1}{(1-x)^2}&=3x^2\cdot\frac{d}{dx}\left[\frac{1}{1-x}\right]\\
&=3x^2\cdot\frac{d}{dx}\left[\sum_{n=0}^\infty x^n\right]\\
&= 3x^2\sum_{n=1}^\infty nx^{n-1}\\
&=\sum_{n=1}^\infty 3nx^{n+1}\\
&=\sum_{n=2}^\infty 3(n-1)x^n
\end{align}$
A: Your work is correct (except that the $3$ should not be squared):
$$\frac{3x^2}{{\left(1-x\right)}^2}=3x^2\left(\sum_{n=0}^{\infty}{x^n}\right)^2\ne3^2x^2\left(\sum_{n=0}^{\infty}{x^n}\right)^2$$
However, an easier way to proceed is to start with:
$$\frac{1}{1-x}=\sum_{n=0}^{\infty}{x^n}$$
and take the derivative of both sides. This gives:
$$\frac{1}{{\left(1-x\right)}^2}=\sum_{n=1}^{\infty}{nx^{n-1}}=\sum_{n=0}^{\infty}{\left(n+1\right)x^n}$$
Can you take it from here?
Edit: Usually it is cumbersome to square a power series. If you carefully multiply a series by itself, you will find that:
$$\left(\sum_{n=0}^{\infty}{a_nx^n}\right)^2=\sum_{n=0}^{\infty}{\left(\sum_{k=0}^{n-1}{a_ka_{n-k}}\right)x^n}$$
Squaring the geometric series is not too complicated, however, since $a_n=1$ for all $n$. You can check for yourself that this formula gives the same result as above.
