Question

Let the $m$ consecutive natural numbers be $n+1,n+2,\dots ,n+m$

Prove that sum of cubes of any number of consecutive natural numbers is always divisible by the sum of these numbers.

Answer 1

Let the consecutive numbers be $(n+1),(n+2),.....(n+m)$

To prove $\frac{\sum _{i=1}^m(n+i{)}^3}{\sum _{i=1}^m(n+1)}$ is an integer.

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

$=\frac{1^3+2^3+...\dots +(n+m{)}^3-(1^3+2^3+.....+n^3)}{(1+2+...\dots +n+m)-(1+2+3+......n)}$

$=\frac{{\left[\frac{(n+m)(n+m+1)}{2}\right]}^2-{\left[\frac{n(n+1)}{2}\right]}^2}{\frac{(n+m)(n+m+1)}{2}-\frac{n(n+1)}{2}}$

$=\frac{(n+m)(n+m+1)}{2}+\frac{n(n+1)}{2}$

Since both pairs we have consecutive numbers, the numerators are even and hence the resulting value is an integer.