Question

PROVE THAT 9*S2^2=S3(1+8*S1) where Sn is sum of n terms in arthematic progression.

Answer 1

$S_{1}$→ Σn = $\frac{n(n+1)}{2}$.
$S_{2}$ → Σn² = $\frac{n(n+1)(2n+1)}{6}$.
$S_{3}$ → Σn³ = $\frac{n^{2}(n+1)^{2}}{4}$.
Consider, L.H.S. = $(3S_{2})^{2}$.
= $(3\frac{n(n+1)(2n+1)}{6})^{2}$
= $(\frac{n(n+1)}{2})^{2}(2n+1)^{2}$
= $S_{3}(1+(4n^{2}+4n))$
= $S_{3}(1+4n(n+1))$
= $S_{3}(1+8 \frac{n(n+1)}{2})$
= $S_{3}(1+8S_{1})$ = R.H.S.