Why is $(n+1)^3 = \sum_{k=0}^n (k+1)^3 - k^3$?

Question

It seems this relation came from some application of the binomial theorem and some combination of the telescope series, but it's not obvious to me how it came about. Can someone explain?

Answer 1

Telescoping series:

$$\sum_{k=0}^n (f(k+1)-f(k)) = f(n+1)-f(0)$$

Answer 2

You’re missing crucial parentheses: the correct expression is

$$(n+1)^3=\sum_{k=0}^n\left((k+1)^3-k^3\right)\,.$$

This is evident if you write out a few examples with small $n$, because the sum telescopes, but it can be demonstrated algebraically as follows:

$$\begin{align*} \sum_{k=0}^n\left((k+1)^3-k^3\right)&=\sum_{k=0}^n(k+1)^3-\sum_{k=0}^nk^3\\ &=\sum_{k=1}^{n+1}k^3-\sum_{k=0}^nk^3\\ &=(n+1)^3+\sum_{k=1}^nk^3-\left(0^3+\sum_{k-1}^nk^3\right)\\ &=(n+1)^3-0^3\\ &=(n+1)^3\,. \end{align*}$$

There’s no need to bring in the binomial theorem.

Answer 3

\begin{align} \sum_{k=0}^n \left((k+1)^3 - k^3\right) &=\sum_{k=0}^n \left(-k^3+(k+1)^3 \right)\\ &=\left(-0^3+1^3\right) + \left(-1^3+2^3\right) + \dots + \left(- n^3+(n+1)^3\right)\\ &=-0^3+(1^3-1^3)+(2^3-2^3) + \dots +(n^3- n^3)+(n+1)^3\\ &=(n+1)^3 \end{align}