Telescoping Series I have a question about a particular formula that is supposed to be used to simplify difficult summations into telescoping series. The formula is as follows. 
$$\sum k(k+1) = \sum \frac{1}{3} \Big(k(k+1)(k+2) - (k-1)k(k+1)\Big)    $$
So now here is the question. How would this be used if one were to sum $k^3$? Thank you very much in advance and I sincerely apologize for any MathJax errors in advance. 
 A: You can usually go about it this way:
$$(k+1)^4 - k^4 =4k^3+6k^2+4k+1$$
Now sum from $k=1$ to $n$
$$\sum_{k=1}^n (k+1)^4 - k^4 =\sum_{k=1}^n4k^3+\sum_{k=1}^n6k^2+\sum_{k=1}^n4k+\sum_{k=1}^n1$$
The LHS telescopes so
$$(n+1)^4-1 =\sum_{k=1}^n4k^3+\sum_{k=1}^n6k^2+\sum_{k=1}^n4k+\sum_{k=1}^n1$$
We know how to compute the other sums, so you have
$$(n+1)^4-1 =\sum_{k=1}^n4k^3+6 \frac{n(2n+1)(n+1)}{6}+4\frac{n(n+1)}{2}+n$$
$$(n+1)^4-1 =\sum_{k=1}^n4k^3+ {n(2n+1)(n+1)}+2{n(n+1)}+n$$
This will ultimately give
$$\frac{{{n^4} + 2{n^3} + {n^2}}}{4} = \sum\limits_{k = 1}^n {{k^3}} $$
Which you can you simplify to
$${\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} = \sum\limits_{k = 1}^n {{k^3}}   $$
A: We have
$$k(k+1)(k+2)= \frac{1}{4}(k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)),$$
which gives a telescoping sum to yield $$\sum_{k=1}^n k(k+1)(k+2)=\frac{1}{4}n(n+1)(n+2)(n+3).$$
Similarly, from the identity you wrote we get $$\sum_{k=1}^n k(k+1)=\frac{1}{3}n(n+1)(n+2),$$ and from the identity $$k=\frac{1}{2}(k(k+1)-(k-1)k)$$ we get $$\sum_{k=1}^n k = \frac{1}{2}n(n+1).$$
Finally, since $$k^3=k(k+1)(k+2) - 3k(k+1) +k,$$ you can combine these results to get 
$$\sum_{k=1}^n k^3 = \frac{1}{4}n(n+1)(n+2)(n+3) -3\cdot\frac{1}{3}n(n+1)(n+2)+\frac{1}{2}n(n+1).$$  This can be written in a nicer form.

Here's a somewhat easier way: $$k^3=(k-1)k(k+1) + k,$$ so a similar telescoping yields
$$\sum_{k=1}^n k^3=\frac{1}{4}(n-1)n(n+1)(n+2)+\frac{1}{2}n(n+1).$$  (Of course both simplify to the same answer Peter shows.)
A: Pedro Tamaroff expands the fourth power of a binomial, but there is a simpler way: only the second power of a binomial need be expanded. Let
\begin{align*}
A(k)&=k^2(k+1)^2\\
\Rightarrow A(k-1)&=k^2(k-1)^2\\
\Rightarrow A(k)-A(k-1)&=k^2[(k+1)^2-(k-1)^2]\\
&=k^2\left[(k^2+2k+1)-(k^2-2k+1)\right]\\
&=4k^3\\
\end{align*}
Sum this last equation for $k=1,\dots,n$. Then
\begin{align*}
4\sum_{k=1}^n k^3&=A(n)\\
&=n^2(n+1)^2\\
\Rightarrow \sum_{k=1}^n k^3&=\dfrac{n^2(n+1)^2}4\\
\end{align*}
