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

(N/A) Let the given statement be $P(n),$ i.e.,
$P(n): 1+3+3^{2}+\ldots+3^{n-1}=\frac{3^{n}-1}{2}$
For $n=1,$ we have
$P(1): 1 = \frac{3^{1}-1}{2} = \frac{2}{2} = 1,$ which is true.
Assume $P(k)$ is true for some positive integer $k,$ i.e.,
$1+3+3^{2}+\ldots+3^{k-1} = \frac{3^{k}-1}{2}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$1+3+3^{2}+\ldots+3^{k-1}+3^{(k+1)-1}$
$= (1+3+3^{2}+\ldots+3^{k-1}) + 3^{k}$
$= \frac{3^{k}-1}{2} + 3^{k}$ [Using $(i)$]
$= \frac{3^{k}-1 + 2 \cdot 3^{k}}{2}$
$= \frac{(1+2) \cdot 3^{k} - 1}{2}$
$= \frac{3 \cdot 3^{k} - 1}{2}$
$= \frac{3^{k+1}-1}{2}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N.$