evaluate convergent power series given
$$\sum_{n=0}^\infty x^{n} (n^{2} + n)$$
so using ratio test I have proven that it converges if and only if
$$|x| < 1$$
but I'm not sure how to evaluate this infinite sum.
so I thought about deriving it, but I'm not sure whether I should do it with respect to 
x or n.
 A: HINT:
$$\dfrac{d(x^n)}{dx}=nx^{n-1}$$
$$\dfrac{d^2(x^n)}{dx^2}=n(n-1)x^{n-2}$$
$$Ax\dfrac{d(x^n)}{dx}+Bx^2\dfrac{d^2(x^n)}{dx^2}=x^n[An+Bn(n-1)]=x^n[Bn^2+n(A-B)]$$
We need $A-B=1,B=1$
Now $\displaystyle\sum_{r=0}^\infty x^r=\dfrac1{1-x}$ for $|x|<1$
A: $$\begin{align}\sum_{n=0}^\infty x^{n} (n^{2} + n) &=\sum_{n=0}^\infty n(n+1)x^n\\~\\&=\sum_{n=0}^\infty x\left(x^{n+1}\right)''\\~\\&= x \left(\sum_{n=0}^\infty x^{n+1}\right)''\\~\\&=x \left(\dfrac{x}{1-x}\right)''\\~\\&=\dfrac{2x}{(1-x)^3}\end{align}$$
A: Let $$S(x)=\sum_{n=0}^\infty x^{n+1}=\frac1{1-x}-1.$$
Then
$$\left(S(x)\right)''=\sum_{n=0}^{\infty}(n+1)nx^{n-1}=\frac2{(1-x)^3},$$
and
$$\sum_{n=0}^{\infty}(n^2+n)x^n=\frac{2x}{(1-x)^3}.$$
A: Hint
Since $n^2=n(n-1)+n$,$$A=\sum_{n=0}^\infty (n^{2} + n)x^{n} =\sum_{n=0}^\infty \big(n(n-1) + 2n\big)x^{n} =\sum_{n=0}^\infty n(n-1)x^{n} +2\sum_{n=0}^\infty nx^{n} $$ which can write $$A=x^2\sum_{n=0}^\infty n(n-1)x^{n-2} +2x\sum_{n=0}^\infty nx^{n-1} $$ You recognize that first term is the second derivative of $B=\sum_{n=0}^\infty x^{n}$ while the second term is the first derivative of the same expression. So $$A=x^2\frac{d^2B}{dx^2}+2x\frac{dB}{dx}=\frac{d}{dx}\Big(x^2\frac{dB}{dx}\Big)$$ while $B=\frac{1}{1-x}$.
I am sure that you can take from here.
Edit
More generally, the trick is to write $$n^2=n(n-1)+n$$ $$n^3=n(n-1)(n-2)+3n(n-1)+n$$ $$n^4=n(n-1)(n-2)(n-3)+6n(n-1)(n-2)+7n(n-1)+n$$  and so on.
