Simplify $\sum n(n+1)$ According to WolframAlpha,
$$\sum_{n=1}^k n(n+1)=\frac{1}{3}k(k+1)(k+2)$$
and it is easy to verify this if we use induction.
However, I would like to know how one can actually come up with this, other than by thinking about how to force the terms to cancel out.
I tried: Since $k(k+1)/2=1+2+\cdots+k$,
$$\begin{align}
\sum_{n=1}^k n(n+1) &= 2\bigg(1(k)+2(k-1)+3(k-2)+\cdots+k(1)\bigg)\\
& = 2\sum_{n=1}^k n(k-n+1)\\
\sum_{n=1}^k n(n+1) &= 2\bigg(1(k)+2(k-1)+3(k-2)+\cdots+k(k-(k-1))\bigg)\\
& = 2\bigg( (1+2+3+\cdots+k)k-(0\cdot1+1\cdot2+2\cdot3+\cdots+(k-1)k) \bigg)\\
& = k^2(k+1)-2\sum_{n=1}^{k-1} n(n+1)\\
& = \sum_{n=1}^k (-2)^{k-n}n^2(n+1)
\end{align}$$
I feel like I'm only making it worse..
 A: Your solution is very close. Your last calculation shows
\begin{equation}
\sum_{n=1}^{k}n(n+1)=k^2(k+1)-2\sum_{n=1}^{k-1}n(n+1).
\end{equation}
The sums on the left and right side are almost the same, except for the coefficient and the limits of summation. If you note
$$\sum_{n=1}^{k-1}n(n+1)=\sum_{n=1}^{k}n(n+1)-k(k+1),$$
then you can immediately write the first identity as
\begin{align}
\sum_{n=1}^{k}n(n+1)&=k^2(k+1)-2\sum_{n=1}^{k}n(n+1)+2k(k+1)\\
&=k(k+1)(k+2)-2\sum_{n=1}^{k}n(n+1).
\end{align}
Now, both sides have your desired sum $\sum_{n=1}^{k}n(n+1)$, which you can solve for and you have your result.
A: A possible way to do it
$$S_n=\sum_{n=1}^k n(n+1) x^n=\sum_{n=1}^k \big[n(n-1)+2n \big]x^n=\sum_{n=1}^k n(n-1) x^n+2\sum_{n=1}^k n x^n$$
$$S_n=x^2\sum_{n=1}^k n(n-1) x^{n-2}+2x\sum_{n=1}^k n x^{n-1}$$
$$S_n=x^2\Big[\sum_{n=1}^k  x^{n}\Big]''+2x\Big[\sum_{n=1}^k  x^{n}\Big]'$$
$$\sum_{n=1}^k  x^{n}=\frac{x \left(x^k-1\right)}{x-1}$$ Compute the derivatives, simplify and look at the limit when $x\to 1$.
A: You can do the sum backwards:
$$\frac{n(n+1)(n+2)}{3}-n(n+1)$$
$$=\frac{(n-1)n(n+1)}{3}$$
etc...
A: If $p$ is a polynomial of degree $d$ and $$P(k)=\sum_{n=1}^kp(n)$$ then $P$ is a polynomial of degree $d+1$. So evaluating $P$ at $d+2$ points gives you $d+2$ equations in the coefficients of $P$.
A simpler example: $p(n)=n$. Say $P(k)=Ak^2+Bk+C$. Then $$A+B+C=P(1)=1,$$ $$4A+2B+C=P(2)=3,$$ and $$9A+3B+C=P(3)=6$$; this leads to $A=B=1/2$, $C=0$, hence $P(n)=n(n+1)/2$.
A: Note that $n(n+1) = 2\bigg(\frac{n(n+1)}{2!}\bigg)=2\binom{n+1}{2}$. Then the sum becomes ....
$$\begin{align}
S(k) &= \sum_{n=1}^k{n(n+1)}\\
&= 2\sum_{n=1}^k{\binom{n+1}{2}}\\
(1) &= 2\binom{k+2}{3}\\
&= 2\cdot\frac{k(k+1)(k+2)}{3!}\\
&= \frac{k(k+1)(k+2)}{3}\\
\end{align}$$
Where $(1)$ is just the hockey-stick identity.
