Let the given statement be P(n), i.e.,
P (n) : 1 · 2 + 2 · 22 + 3 · 23 + ... + n · 2n = (n–1)2n+1 + 2.
First we prove that the statement is true for n = 1.
P (1) : 1⋅2 = (1 – 1) 21+1 + 2 ⇒ 2 = 0 + 2 = 2, which is true.
Assume, P(k) is true for some positive integer k, i.e.,
1 · 2 + 2 · 22 + 3 · 23 + ... + k · 2k = (k–1)2k+1 + 2. ... (i)
We shall now prove that P(k + 1) is also true.
for this we have to prove that
1 · 2 + 2 · 22 + 3 · 23 + ... + k · 2k + (k + 1) . 2k+1 = (k + 1 - 1) 2k+1+1 + 2
L.H.S. = 1 · 2 + 2 · 22 + 3 · 23 + ... + k · 2k + (k + 1) . 2k+1 = (k - 1) 2k+1 + 2 + (k + 1) . 2k+1
= 2k+1 (k - 1 + k + 1) + 2 = 2k+1 . 2k + 2 = 2k+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.