Let the given statement be P(n), i.e., P (n) : $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + ... + $\frac{1}{(3n-2)(3n+1)}$ = $\frac{n}{3n+1}$ First we prove that the statement is true for n = 1. P (1) : $\frac{1}{1.4}$ = $\frac{1}{3.1+1}$ ⇒ $\frac{1}{4}$ = $\frac{1}{4}$ , which is true Assume P(k) is true for some positive integer k, i.e., $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + ... + $\frac{1}{(3k-2)(3k+1)}$ = $\frac{k}{3k+1}$ ... (i) Now we shall prove that P(k + 1) is also true. For this we have to prove that $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + ... + $\frac{1}{(3k-2)(3k+1)}$ + $\frac{1}{(3(k+1)-2)(3(k+1)+1)}$ = $\frac{k+1}{3(k+1)+1}$ L.H.S. = $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + ... + $\frac{1}{(3k-2)(3k+1)}$ + $\frac{1}{(3(k+1)-2)(3(k+1)+1)}$ = $\frac{k}{3k+1}+\frac{1}{(3k+1)(3k+4)}$ [From (i)] = $\frac{1}{3k+1}\left[k+\frac{1}{3k+4}\right]$ = $\frac{1}{3k+1}\left[\frac{3k^2+4k+1}{3k+4}\right]$ = $\frac{(3k+1)(k+1)}{(3k+1)(3k+4)}$ = $\frac{k+1}{3k+4}$ = $\frac{k+1}{3(k+1)+1}$ = R.H.S Thus P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true ∀ n ∈ N.