Summation formula for $x^2+x$ Since I learned easier ways of calculating summations I've been curious as to how I could find formulas for as many equations as possible. I came across the equation $x^2+x$, I've spent quite some time on this problem and could not find a solution. If someone has maybe already done this or have any suggestions on how I could get the formula that would be greatly appreciated. 
Example of another summation with a equation:
$\sum\limits_{i=1}^n$  =   $x^2$
Equation to solve this is $\frac{n(n+1)(2n+1)}{6}$  
 A: Extending what Git Gud said:
$$\sum^n_{k=0}\left(k^2+k\right)=\sum^n_{k=0}k^2+\sum^n_{k=0}k=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n^3}{3}+n^2+\frac{2n}{3}$$
A: Note that 
$$k^2+k=\frac{1}{3}\left((k+1)^3-k^3-1\right).$$
Thus our sum $\sum_{k=1}^n (k^2+k)$ is equal to
$$\frac{1}{3}\left((2^3-1^3-1)+(3^3-2^3-1)+(4^3-3^3-1)+(5^3-4^3-1)+\cdots +((n+1)^3-n^3-1)\right).$$
Observe the nice almost total cancellations. We end up with
$$\frac{1}{3}\left((n+1)^3-1^3-n\right).$$
Remarks: $1.$ Since we know $\sum_1^n k$, this gives a way to derive the formula for $\sum_1^n k^2$. 
$2.$ The sums $\sum k(k+1)$, $\sum k(k+1)(k+2)$, $\sum k(k+1)(k+2)(k+3)$ and so on are nice, much nicer than $\sum k^2$, $\sum k^3$, $\sum k^4$ and so on. 
A: Note that $$\frac{n^2+n}{2} = {n+1 \choose 2}.$$  Then by induction $$\begin{eqnarray}{n+2 \choose 3}&=&{n+1 \choose 2}+{n+1 \choose 3}\\
&=&{n+1 \choose 2} + \sum_{k=1}^{n-1}{k+1 \choose 2}\\
&=& \sum_{k=1}^{n}{k+1 \choose 2}.
\end{eqnarray}$$ So $$\sum_{k=1}^{n}(k^2+k)=2{n+2 \choose 3}.$$ This directly generalizes to binomial sums $$\sum_{k=1}^n{k + m-1 \choose m}.$$ This is particularly obvious if you draw what is going on in Pascal's triangle.
