Question

Prove that : $1+3+3^2+...+3^{n-1}=\frac{(3^n-1)}{2}$

Answer 1

The given series $1+3+3^2+......+3^{n-1}$ is in G.P. where

first term, $a_1=1$

common ratio, $r=\frac{a_2}{a_1}=\frac{3}{1}=3$

$n$th term, $a_n=a_1{r}^{n-1}=3^{n-1}$

Sum of $n$ terms,

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

$=\frac{1(3^n-1)}{3-1}$

$=\frac{3^n-1}{2}$