sequences and series - Finding formula for partial sum of polynomial terms?

For example, we know that the following is true (and can be easily derived):

$\sum\limits_{x=1}^{n}x = \frac{1}{2}n(n+1)$

But, what if we wanted to find the sum of a series like this:

$\sum\limits_{x=1}^{n}x(2x+1)$

Wolfram|Alpha tells me that the answer is $\frac{1}{6}n(n+1)(4n+5)$, but I'm at a loss as to how it came up with this answer. Is there a simple method for finding the general formula for a partial sum of the form $\sum\limits_{x=1}^{n}y(x)$ where $y(x)$ is a polynomial with rational roots?

Answer

Let $p(x)=\sum\limits_{i=0}^{n} a_{i}x^{i}$

Say you wan't to find $\sum p(x)$.

There is a general formula for $\sum x^{k}$.

$$1^{k}+2^{k}+\cdots+n^{k}=\sum\limits_{i=1}^{k}S(k,i)\binom{n+1}{i+1}i!\\=\frac{n^{k+1}}{k+1}+\frac{1}{2}n^{k}+B_{1}\frac{r}{2!}n^{r-1}-\cdots$$

Where $S(k,i)$ is the Stirling number of the Second Kind and $B_{r}$ denots the the Bernoulli's Numbers.

So using that you can find the sum of any polynomial.

The given sum- $$\sum x(2x+1)=2\sum x^{2}+\sum x=2\binom{n+1}{2}+4\binom{n+1}{3}+\binom{n+1}{2}=\frac{1}{6}n(n+1)(4n+5)$$