Let the given statement be P(n), i.e., 1 . 3 + 2 . $3^2$ + 3 . $3^3$ + ... + $n \cdot 3^n$ = $\frac{(2n-1)3^{n+1}+3}{4}$ First we prove that the statement is true for n = 1. P (n) : 1 . 3 = $\frac{(2.1-1)3^{1+1}+3}{4}$ ⇒ 3 = $\frac{3^2+3}{4}$ = $\frac{9+3}{4}$ = $\frac{12}{4}$ = 3 which is true. Assume P(k) is true for some positive integer k, i.e., 1 . 3 + 2 . $3^2$ + 3 . $3^3$ + ... + $k \cdot 3^k$ = $\frac{(2k-1)3^{k+1}+3}{4}$ ... (i) We shall now prove that P(k + 1) is also true. For this we have to prove that 1 . 3 + 2 . $3^2$ + 3 . $3^3$ + ... + $k \cdot 3^k + (k+1) \cdot 3^{k+1}$ = $\frac{[2(k+1)-1]3^{k+2}+3}{4}$ L.H.S. = 1 . 3 + 2 . $3^2$ + 3 . $3^3$ + ... + $k \cdot 3^k + (k+1) \cdot 3^{k+1}$ = $\frac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}$ [From (i)] = $\frac{(2k-1+4k+4)3^{k+1}+3}{4}$ = $\frac{(6k+3)3^{k+1}+3}{4}$ = $\frac{[2(k+1)-1]3^{k+1+1}+3}{4}$ = R.H.S. Thus P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.