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) < k2 + 6k + 9 + 2 = k2 + 6k + 11 < k2 + 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.