Question

Find the sum $S_n$ of the cubes of the first n terms of an A.P. and show that the sum of first n terms of the A.P. is a factor of $S_n$

Answer 1

Let the A.P. be$(a+d)+(a+2d)+....+(a+nd)$
$p_n=Sum=\frac{n}{2}\left[\right(a+d)+(a+d\left)\right]=\frac{n}{2}[2a+(n+1\left)d\right]...\left(1\right)$
$S_n=(a+d{)}^3+(a+2d{)}^3+....+(a+nd{)}^3$
$=n{a}^3+3a^{3}d\sum n+3a{d}^2\sum n^2+d^3\sum n^3$
$=n{a}^3+3a^{3}d\cdot \frac{n(n-1)}{2}+3a{d}^2\cdot \frac{n(n+1)(2n+1)}{6}+d^3\cdot \frac{n^2(n+1{)}^2}{4}$
$=\frac{n}{4}[4a^3+6a{d}^2(n+1)+2a{d}^2(n+1\left)\right(2n+1)+d^{3}n(n+1{)}^2]$
$\frac{1}{2}\cdot \frac{n}{2}[2a+(n+1\left)d\right][2a^2+2ad(n+1)+d^{2}n(n+1\left)\right]......\left(2\right)$
$=\frac{1}{2}{P}_n[2a^2+2ad(n+1)+d^{2}n(n+1\left)\right]by\left(1\right).....\left(3\right)$
$=P_n[a^2+ad(n+1)+d^2\lambda ]$
$\therefore n(n+1)=Productoftwoconsecutivenumberiseven=2\lambda ,say$
By actual division (2) given $S_n$ and (3) shows that $P_n$ is a factor of $S_n$.