Let the given statement be P(n), P (n) : (2n + 7) < $(n+3)^2$ First we prove that the statement is true for n = 1. P(1) : (2·1 + 7) < $(1 + 3)^2$ i.e., 9 < 16, which is true. Assume P(k) is true. i.e., (2k + 7) < $(k + 3)^2$ ...(i) Now prove that P(k + 1) is true For this we have to prove that [2(k + 1) + 7] < $[(k + 1) + 3]^2$ Since (2k + 7) + 2 < $(k + 3)^2$ + 2 [From (i)] ⇒ (2k + 2 + 7) < $k^2$ + 6k + 9 + 2 = $k^2$ + 6k + 11 < $k^2$ + 8k + 16 = $(k + 4)^2$ = $(k + 1 + 3)^2$ ⇒ 2(k + 1) + 7 < $(k + 1 + 3)^2$ Thus P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true ∀ n ∈ N.