Let the given statement be P(n), i.e., $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}$ + ... + $\frac{1}{(2n+1)(2n+3)}$ = $\frac{n}{3(2n+3)}$ First we prove that the statement is true for n = 1. P (1) : $\frac{1}{3.5}$ = $\frac{1}{3(2.1+3)}$ ⇒ $\frac{1}{15}$ = $\frac{1}{15}$ , which is true Assume P(k) is true for some positive integer k, i.e., L.H.S. = $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}$ + ... + $\frac{1}{(2k+1)(2k+3)}$ = $\frac{k}{3(2k+3)}$ ... (i) Now we shall prove that P(k + 1) is true. For this we have to prove that $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}$ + ... + $\frac{1}{(2k+1)(2k+3)}$ + $\frac{1}{(2(k+1)+1)(2(k+1)+3)}$ = $\frac{k+1}{3\left[2(k+1)+3\right]}$ L.H.S. = $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}$ + ... + $\frac{1}{(2k+1)(2k+3)}$ + $\frac{1}{(2(k+1)+1)(2(k+1)+3)}$ = $\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)}$ [From (i)] = $\frac{1}{2k+3}\left[\frac{k}{3}+\frac{1}{2k+5}\right]$ = $\frac{1}{2k+3}\left[\frac{2k^2+5k+3}{3(2k+5)}\right]$ = $\frac{(2k+3)(k+1)}{3(2k+3)(2k+5)}$ = $\frac{k+1}{3(2k+5)}$ = $\frac{k+1}{3\left[2(k+1)+3\right]}$ = R.H.S. Since k ≠ -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.