Let the given statement be P(n), i.e., P (n) : $\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}$ + ... + $\frac{1}{(n+1)(n+2)}$ = $\frac{n(n+3)}{4(n+1)(n+2)}$ First we prove that statement is true for n = 1. P (1) : $\frac{1}{1\cdot2\cdot3}$ = $\frac{1\cdot(1+3)}{4(1+1)(1+2)}$ ⇒ $\frac{1}{6}$ = $\frac{4}{4\cdot2\cdot3}$ = $\frac{1}{6}$ which is true Assume P(k) is true for some positive integer k, i.e., $\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}$ + ... + $\frac{1}{(k+1)(k+2)}$ = $\frac{k(k+3)}{4(k+1)(k+2)}$ ... (i) We shall now prove that P(k + 1) is also true. For this we have to prove that $\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}$ + ... + $\frac{1}{(k+1)(k+2)}$ + $\frac{1}{(k+1)(k+2)(k+3)}$ = $\frac{(k+1)(9k+4)}{4(k+2)(k+3)}$ L.H.S. = $\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}$ + ... + $\frac{1}{(k+1)(k+2)}$ + $\frac{1}{(k+1)(k+2)(k+3)}$ = $\frac{k(k+3)}{4(k+1)(k+2)}$ + $\frac{1}{(k+1)(k+2)(k+3)}$ [From (i)] = $\frac{1}{(k+1)(k+2)}$ $\left[\frac{k(k+3)}{4}+\frac{1}{k+3}\right]$ = $\frac{1}{(k+1)(k+2)}$ $\left[\frac{k(k+3)^2+4}{4(k+3)}\right]$ = $\frac{k(k^2+6k+9)+4}{4(k+1)(k+2)(k+3)}$ = $\frac{k^3+6k^2+9k+4}{4(k+1)(k+2)(k+3)}$ = $\frac{(k+1)(k^2+5k+4)}{4(k+1)(k+2)(k+3)}$ = $\frac{(k+1)(k+1)(k+4)}{4(k+1)(k+2)(k+3)}$ = $\frac{(k+1)(k+2)}{4(k+2)(k+3)}$ Thus, P(k + 1) is true, whenever P(k) is true. Hence, by the principal of mathematical induction P(n) is true.