Question

Find the sum of the cubes of the terms of an A.P., and show that it is exactly divisible by the sum of the terms.

Answer 1

$S=\sum (a+(n-1)d{)}^3$

$S=n{a}^3+\frac{3a^{2}d(n-1)n}{2}+\frac{3a{d}^2(n-1)n(2n-1)}{6}+\frac{d^3(n-1{)}^2{n}^2}{4}$

$S=\frac{n}{4}\{4a^3+6a^{2}d(n-1)+2a{d}^2(n-1\left)\right(2n-1)+d^{3}n(n-1{)}^2\}$

$S=\frac{n}{4}(2a+(n-1\left)d\right)\{2a^2+2(n-1)ad+n(n-1\left)d^2\right\}$

$S=\frac{n}{2}(2a+(n-1\left)d\right)\{a^2+(n-1)ad+\frac{n(n-1)}{2}{d}^2\}$

$\frac{n(n-1)}{2}$ is an integer since product of 2 consecutive numbers is always divisible by $2$

Hence, $S$ is exactly divisible by the sum of the A.P., which is $\frac{n}{2}(2a+(n-1\left)d\right)$