What is the sum of $\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$? Consider the power sequence
$$\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$$
What is the function to which it sums to?
My reasoning is to differntiate the sum with respect to $x$, then to integrate with respect to x from $0$ to $x$ after variation of the sum into a Taylor series form.
I mean:
$$\int _0 \frac{d}{dx}\sum_{n=1}^\infty (n^2+n^3)x^{n-1}dx=\int_0 \sum_{n=1}^\infty \frac{n^2+n^3}{n-1}x^{n-2}dx$$
But it doesn't seem to work in this exercise.

This is a very general approach.
$$n^3+n^2 = n(n-1)(n-2) + 4n(n-1) + 2n$$
So $$(n^3+n^2)x^n = \left(x^3\left(\frac{d}{dx}\right)^3 + 4x^2\left(\frac{d}{dx}\right)^2 + 2x\frac{d}{dx}\right)x^n$$
So $$\sum (n^3+n^2)x^n =\left(x^3\left(\frac{d}{dx}\right)^3 + 4x^2\left(\frac{d}{dx}\right)^2 + 2x\frac{d}{dx}\right)\frac{1}{1-x}$$
In general, if $(n)_k = n(n-1)\cdots(n-k+1)$ is the falling factorial, then these are a basis for all polynomials, so if $p(n)$ is a polynomial of degree $d$, then we can write:
$$p(n)=\sum_{k=0}^{d} a_k(n)_k$$
Then $$\sum_{n=0}^\infty p(n)x^n = \left(\sum_{k=0}^d a_kx^k\left(\frac{d}{dx}\right)^k\right)\frac{1}{1-z}$$
Now $$\left(\frac{d}{dx}\right)^k\frac{1}{1-x} = \frac{k!}{(1-x)^{k+1}}$$
So that gives:
$$\sum_{n=0}^\infty p(n)x^n = \sum_{k=0}^d \frac{k!a_kx^k}{(1-x)^{k+1}}=\frac{\sum_{k=0}^d k!a_kx^k(1-x)^{d-k}}{(1-x)^{d+1}}$$
This shows that $(1-x)^{d+1}\sum_{n=0}^\infty p(n)x^n$ is a polynomial of degree at most $d$. The beauty of this is that you can just multiply polynomials:
$$(1-x)^{d+1}\sum_{n=0}^d p(n)x^n$$ and ignore the terms of degree bigger than $d$ to figure out what the numerator polynomial is.
So if $p(n)=n^3+n^2$ then 
$$\begin{align}(1-x)^4(2x+12x^2+36x^3) &= 2x(1-4x+6x^2-4x^3+x^4)(1+6x+18x^2) \\
&= 2x(1+2x +0x^2 +\dots)
\end{align}$$
And your series is:
$$\sum_{n=1}^\infty (n^3+n^2)x^{n-1} = \frac{1}{x}\sum_{n=0}^\infty p(n)x^n=\frac{2(1+2x)}{(1-x)^4}$$
Interesting to note that $(1-x)^4\sum_{n=0}^d p(n)x^d$ is essentially the same as the approach in Semiclassical's answer - multiplying by $(1-x)$ repeatedly is essentially the same as doing his finite differences.
