Converting a power series to a function 
$$\sum (n+1)^2x^n = {1+x\over (1-x)^3}$$

I tried to find a similar power series and start from there. I see $ x^n $ and it reminds me of geometric series, but $(n+1)^2$ confuses me. How do I prove this?
 A: The geometric series:
$$\sum_{n=0}^\infty x^n=\frac1{1-x}$$
Multiply both sides by $x$.
$$\sum_{n=0}^\infty x^{n+1}=\frac x{1-x}$$
Take the derivative of both sides:
$$\sum_{n=0}^\infty(n+1)x^n=\frac d{dx}\frac x{1-x}$$
Multiply both sides by $x$.
$$\sum_{n=0}^\infty(n+1)x^{n+1}=x\cdot\frac d{dx}\frac x{1-x}$$
Take the derivative of both sides:
$$\sum_{n=0}^\infty(n+1)^2x^n=\frac d{dx}\left(x\cdot\frac d{dx}\frac x{1-x}\right)$$
A: First use the fact that
$$\sum_{n=0}^\infty x^{n+1} = {x\over 1-x}$$
now take the derivative to get
$$\sum_{n=0}^\infty (n+1)x^n = {1\over (1-x)^2}$$
Now multiply by $x$ and take the derivative again:
$$\sum_{n=0}^\infty (n+1)^2x^n = {d\over dx}\left({x\over (1-x)^2}\right) = {(1-x)^2+2x(1-x)\over (1-x)^3}$$

$$={1-x^2\over (1-x)^4}={1+x\over (1-x)^3}$$


Another way:  Note that $\displaystyle n^2= {n\choose 2}+{n+1\choose 2}$ is a sum of consecutive triangular numbers and that if $\displaystyle f(x) = \sum a_nx^n$ then $\displaystyle {f(x)\over 1-x}=\sum_k\left(\sum_{m\le k} a_m\right)x^k$. Applying this twice to $\displaystyle f(x) = {1\over 1-x}$ we see

$$G(x) = \sum_n {n+2\choose 2}x^n = {1\over (1-x)^3}$$

Which leads us to

$${1+x\over (1-x)^3}= G(x) + xG(x) = \sum {n+2\choose 2}x^n+{n+2\choose 2}x^{n+1}$$
  $$=\sum\left({n+2\choose 2} + {n+1\choose 2}\right)x^n=\sum (n+1)^2x^n$$

A: Just another way to do it.
$$A=\sum_{n=0}^\infty (n+1)^2x^n =\sum_{n=0}^\infty (n^2+2n+1)x^n $$ Now, write $$n^2+2n+1=(n(n-1)+n)+2n+1=n(n-1)+3n+1$$ $$A=\sum_{n=0}^\infty n(n-1)x^n+3\sum_{n=0}^\infty nx^n+\sum_{n=0}^\infty x^n$$ $$A=x^2\sum_{n=0}^\infty n(n-1)x^{n-2}+3x\sum_{n=0}^\infty nx^{n-1}+\sum_{n=0}^\infty x^n$$ $$A=x^2\left(\sum_{n=0}^\infty x^n \right)''+3x\left(\sum_{n=0}^\infty x^n \right)'+\left(\sum_{n=0}^\infty x^n \right)$$ $$\left(\sum_{n=0}^\infty x^n \right)=\frac{1}{1-x}\implies\left(\sum_{n=0}^\infty x^n \right)'=\frac{1}{(1-x)^2}\implies\left(\sum_{n=0}^\infty x^n \right)''=\frac{2}{(1-x)^3}$$ $$A=\frac{2x^2}{(1-x)^3}+\frac{3x}{(1-x)^2}+\frac{1}{1-x}=\frac{1+x}{(1-x)^3}$$
Using the same steps, you could make the problem more general and show that $$B=\sum_{n=0}^\infty (an^2+bn+c)x^n =\frac{(a-b+c)x^2 + (a+b-2 c)x+c }{(1-x)^3}$$
