Proof for consecutive integers

Question

Prove that if $n$ is an odd integer, $n^3$ is the sum of $n$ consecutive integers.

I'm confused on how to prove something with consecutive integers.

Answer 1

$$1^3=1 \\ 3^3=8+9+10 \\ 5^3=23+24+25+26+27 \\ 7^3=46+47+48+49+50+51+52\\\vdots\\(2n+1)^3=\sum_{i=(2n+1)^2-n}^{(2n+1)^2+n}i=\sum_{i=1}^{(2n+1)^2+n}i-\sum_{i=1}^{(2n+1)^2-n-1}i=\\\frac{\left((2n+1)^2+n\right)\left((2n+1)^2+n+1\right)-\left((2n+1)^2-n\right)\left((2n+1)^2-n-1\right)}{2}.$$ So we have $$\require{cancel}2(2n+1)^3=\cancel{(2n+1)^4}+n(2n+1)^2+(n+1)(2n+1)^2+\cancel{n(n+1)}-\cancel{(2n+1)^4}+\\(n+1)(2n+1)^2+n(2n+1)^2-\cancel{n(n+1)}\\ 2(2n+1)^3=2(2n+1)^3.$$

Answer 2

A little simpler:

$\sum_{i=(2n+1)^2-n}^{(2n+1)^2+n}i =\sum_{i=-n}^{n}((2n+1)^2+i) =(2n+1)^3+\sum_{i=-n}^{n}i =(2n+1)^3 $