Let the given statement be P(n), i.e., P (n) : $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}$ + ... + $\frac{1}{(3n-1)(3n+2)}$ = $\frac{n}{6n+4}$ First we prove that statement is true for n = 1. P (1) : $\frac{1}{2.5}$ = $\frac{1}{6.1+4}$ ⇒ $\frac{1}{10}$ = $\frac{1}{10}$ , which is true Assume P(k) is true for some positive integer k, i.e., $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}$ + ... + $\frac{1}{(3k-1)(3k+2)}$ = $\frac{k}{6k+4}$ (i) We shall now prove that P(k + 1) is also true. For this we have to prove that $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}$ + ... + $\frac{1}{(3k-1)(3k+2)}$ + $\frac{1}{(3k+2)(3k+5)}$ = $\frac{k+1}{6(k+1)+4}$ L.H.S. = $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}$ + ... + $\frac{1}{(3k-1)(3k+2)}$ + $\frac{1}{(3k+2)(3k+5)}$ = $\frac{k}{6k+4} + \frac{1}{(3k+2)(3k+5)}$ [From (i)] = $\frac{1}{3k+2}\left[\frac{k}{2}+\frac{1}{3k+5}\right]$ = $\frac{1}{3k+2}\left[\frac{3k^2+5k+2}{2(3k+5)}\right]$ = $\frac{3k^2+5k+2}{2(3k+2)(3k+5)}$ = $\frac{(3k+2)(k+1)}{2(3k+2)(3k+5)}$ = $\frac{k+1}{6k+10}$ = $\frac{k+1}{6(k+1)+4}$ = R.H.S. Since k ≠ -2/3 Thus P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true ∀ n ∈ N.