Question

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

Answer 1

Let
$P\left(n\right)=1+3+3^2+\cdots +3^{n-1}=\frac{3^n-1}{2}\text{For n = 1}P\left(1\right)=1=\frac{3^1-1}{2}\Rightarrow =1\therefore P\left(1\right)$ is true
Let P (n) be true for n = k
$P(k+1)=1+3+3^2+\cdots +3^{k-1}+3^k=\frac{(3^k-1)}{2}+3^k=\frac{3^k-1+{2.3}^k}{2}=\frac{{3.3}^k-1}{2}=\frac{3^{k+1}-1}{2}$
$\therefore P(k+1)$ is true.
Thus P(k) is true \Rightarrow (k + 1) is true.
Hene by principle of mathematical induction,
P(n) is true for all $nϵN.$