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.

-

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

$${\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} = \sum\limits_{k = 1}^n {{k^3}}$$

-
Incredible answer and very well written. Thank you very much! – Astephen2 Feb 15 '12 at 16:37
@Astephen2 Thanks! You can accept it by clicking in the green tick, top left. – Pedro Tamaroff Feb 15 '12 at 20:01

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.)

-