Let the given statement be P(n), i.e., P (n) : 1 + 2 + 3 + ... + n < $\frac{1}{8}(2n+1)^2$ First we prove that the statement is true for n = 1. P (1) : 1 < $\frac{1}{8}(2.1+1)^2$ = $\frac{1}{8}(3)^2$ = $\frac{9}{8}$ ⇒ 1 < $\frac{9}{8}$ , which is true Assume P(k) is true for some positive integer k, i.e., 1 + 2 + 3 + ... + k < $\frac{1}{8}(2k+1)^2$ ... (i) Now we shall prove that P(k + 1) is true. For this we have to prove that 1 + 2 + 3 + ... + k + (k + 1) < $\frac{1}{8}[2(k+1)+1]^2$ L.H.S. = 1 + 2 + 3 + ... + k + (k + 1) < $\frac{1}{8}(2k+1)^2$ + (k + 1) [From (i)] = $\frac{4k^2+4k+1+8k+8}{8}$ = $\frac{4k^2+12k+9}{8}$ = $\frac{(2k+3)^2}{8}$ = $\frac{[2(k+1)+1]^2}{8}$ = R.H.S. ⇒ 1 + 2 + 3 + ... + k + (k + 1) < $\frac{1}{8}[2(k+1)+1]^2$ Thus P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true ∀ n ∈ N.