<s>[INST] 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. 
 [/INST] 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.)
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