Let the given statement be P(n) i.e., P (n) : 1 + 3 + $3^2$ + ... + $3^{n-1}$ = $\frac{(3^n-1)}{2}$ First we prove that the statement is true for n = 1 P (1) : 1 = $\frac{3^1-1}{2}$ = $\frac{2}{2}$ = 1 , which is true Assume P(k) is true for some positive integers k, i.e., 1 + 3 + $3^2$ + ... + $3^{k-1}$ = $\frac{3k-1}{2}$ ... (i) We shall now prove that P(k + 1) is also true. For this we have to prove that 1 + 3 + $3^2$ + ... + $3^{k-1} + 3^k$ = $\frac{(3^{k+1}-1)}{2}$ By adding $3^k$ to both the sides of (i), we get L.H.S. = 1 + 3 + $3^2$ + ... + $3^{k-1} + 3^k$ = $\frac{3^k-1}{2} + 3^k$ [from (i)] = $\frac{3^k-1+2 \cdot 3^k}{2}$ = $\frac{3 \cdot 3^k - 1}{2}$ = $\frac{3^{k+1}-1}{2}$ = R.H.S. Thus P(k + 1) is true, whenever P(k) is true. Hence, by the principal of mathematical induction, the statement P(n) is true for all n ∈ N.