Question

If $S_1,S_2$ and $S$ denotes respectively the sum of $n$, $2n$ and infinite number of terms of a G.P. then prove that $S_1(S_1-S)=(S_1-S_2)S.$

Answer 1

$S=\frac{a(1-r^n)}{1-r}$

$S_1=\frac{a(1-r^n)}{1-r}$

$S_2=\frac{a(1-r^{2n})}{1-r}$

$S=\frac{a}{1-r}$

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

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

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

$L.H.S$

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

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

$⟹\frac{a^2(2-{2r}^n)}{{(1-r)}^2}$

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

$\therefore R.H.S=L.H.S$ $\left(proved\right)$