# Prove the following by using the principle of mathematical induction for all $n \in N$ :  $1+3+3^2+...+3^{n-1}= \large\frac{(3^n-1)}{2}$

Let the given statement be $P(n)$, i.e.,
For $n=1$ we have,
$P(1):1 = \large\frac{(3^1-1)}{2}$$=\large\frac{3-1}{2}$$=\large\frac{2}{2}$$=1, which is true Let P(k) be true for some positive integer k, i.e., 1+3+3^2+...+3^{k-1}= \large\frac{(3^k-1)}{2}-----------(i) We shall now prove that P(k+1) is true. Consider 1+3+3^2+...+3^{k-1}+3^{(k+1)-1} (1+3+3^2+...+3^{k-1})+3^k = \large\frac{(3^k-1)}{2}$$ +3^k \qquad$ [ Using (i) ]
$= \large\frac{(3^k-1)+2.3^k}{2}$
$= \large\frac{(1+2)3^k-1}{2}$
$= \large\frac{3.3^k-1}{2}$
$= \large\frac{3^{k+1}-1}{2}$
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.