Let the given statement be P(n), i.e., P (n) : 1 · 2 + 2 · $2^2$ + 3 · $2^3$ + ... + n · $2^n$ = $(n – 1)2^{n + 1}$ + 2. First we prove that the statement is true for n = 1. P (1) : 1⋅2 = (1 – 1) $2^{1 + 1}$ + 2 ⇒ 2 = 0 + 2 = 2, which is true. Assume, P(k) is true for some positive integer k, i.e., 1 · 2 + 2 · $2^2$ + 3 · $2^3$ + ... + k · $2^k$ = $(k – 1)2^{k + 1}$ + 2. ... (i) We shall now prove that P(k + 1) is also true. for this we have to prove that 1 · 2 + 2 · $2^2$ + 3 · $2^3$ + ... + k · $2^k$ + (k + 1) . $2^{k+1}$ = (k + 1 - 1) $2^{k+1+1}$ + 2 L.H.S. = 1 · 2 + 2 · $2^2$ + 3 · $2^3$ + ... + k · $2^k$ + (k + 1) . $2^{k+1}$ = (k - 1) $2^{k+1}$ + 2 + (k + 1) . $2^{k+1}$ = $2^{k+1}$ (k - 1 + k + 1) + 2 = $2^{k+1}$ . 2k + 2 = $2^{k+1+1}$ (k + 1 - 1) + 2 = R.H.S. Thus, P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true ∀ n ∈ N.