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
$S_1+S_2+S_3+...+S_n=\frac{na}{1-r}-\frac{ar(1-r^n)}{{(1-r)}^2}$

Answer 1

$S_n=\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}-a.\frac{r^n}{1-r}$
Putting $n=1,2,3,..,n$ and adding, we get
$S_1+S_2+S_3+..+S_n$
$=n.\left(\frac{a}{1-r}\right)-\frac{a}{1-r}(r+r^2+r^3+.....+r^n)$
$=\frac{na}{1-r}-\frac{a}{1-r}.r.\frac{(1-r^n)}{1-r}$
$=\frac{na}{1-r}-\frac{ar}{{(1-r)}^2}(1-r^n)$.