Question

If $S_1,S_2,S_3$ be respectively the sums of n, 2n, 3n terms of a G.p., then prove $S_1^2+S_2^2=S_1(S_2+S_3)$.

Answer 1

$S_1=$ sum of n terms,

$S_2=$ sum of 2n terms,

$S_3=$ sum of 3n terms,

Then $S_1^2+S_2^2$

$=(S_n{)}^2+(S_{2n}{)}^2$

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

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

$=\frac{a^2}{(1-r{)}^2}[1+r^{2n}-2r^n+1+r^{4n}-2r^{2n}]$

$=\frac{a^2}{(1-r{)}^2[2-r^{2n}-2r^n+r^{4}n]}$ .....(i)

Also,

$S_1(S_2+S_3)$

$=\frac{a(1-r^n)}{1-r}(\frac{a(1-r^{2n})}{1-r}+\frac{a(1-r^{3n})}{1-r})$

$=\frac{a^2}{(1-r{)}^2}\left[\right(1-r{)}^n(1-r^{2n})+(1-r^n)(1-r^{3n})]$

$=\frac{a^2}{(1-r{)}^2[1-r^{2n}-r^n+r^{3n}-r^{3n}-r^n+1+r^{3n}]}$

$=\frac{a^2}{(1-r{)}^2[1-r^{2n}-r^n+r^{3n}-r^{3n}-r^n+1+r^{4n}]}$

$=\frac{a^2}{(1-r{)}^2}[2-r^{2}n-2r^n+r^{4n}]$ ..... (ii)

$\therefore$ Eq. (i) = Eq. (ii)

Hence, $S_1^2+S_2^2=S_1(S_2+S_3)$