Question

If $S_n$ represents the sum of $n$ terms of a $G.P.$ whose first term and common ratio are $a$ and $r$ respectively, then prove that
$S_1+S_3+S_5+...+S_{2n-1}=\frac{an}{1-r}-\frac{ar(1-r^{2n})}{{(1-r)}^2(1+r)}$

Answer 1

$S_1+S_3+S_5+..+S_{2n-1}$ i.e. $n$ terms Putting $n=1,3,5,..$ in $\left(1\right)$ and adding,
$=n.\frac{a}{1-r}-\frac{a}{1-r}[r+r^3+r^5+.....+r^{2n-1}]$
$=n.\frac{a}{1-r}-\frac{a}{1-r}.\frac{r(1-r^{2n})}{1-r^2}$
$=\frac{na}{1-r}-\frac{ar}{{(1-r)}^2(1+r)}.(1-r^{2n})$