Finding a Suitable Telescoping Sum to Use in Problem

Question

I am currently lost on the following problem:

Use a telescoping sum to give a proof without induction that for each $n \in \mathbb{N},$

$$1^3+2^3+3^3+\dots +n^3=\frac{n^2(n+1)^2}{4}$$

I have followed other examples where they show equalities like this, but I don't understand how they seem to come up with the telescoping series they use to solve them.

Thank you!

Answer 1

For all $k \in \mathbb{N}$, you have $$(k+1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1$$

So $$(k+1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1$$

Summing for $k=1$ to $n$, you get $$(n+1)^4-1=\sum_{k=1}^n (4k^3 + 6k^2 + 4k + 1)$$

So using the well-known values of $\sum_{k=1}^n k^2$ and $\sum_{k=1}^n k$, $$\sum_{k=1}^n k^3 = \frac{1}{4}((n+1)^4-1 - n(n+1)(2n+1) - 2n(n+1) - n)$$

i.e. $$\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$$

(If the values of $\sum_{k=1}^n k^2$ and $\sum_{k=1}^n k$ are not "well-known", you can actually compute them with the same process)

Answer 2

Using the identity $$ k^3=\frac14\bigg[k^2(k+1)^2-(k-1)^2k^2\bigg] $$ it is easy to see $$ \sum_{k=1}^n k^3=\frac14\sum_{k=1}^n\bigg[k^2(k+1)^2-(k-1)^2k^2\bigg]=\frac14n^2(n+1)^2. $$