Let the given statement be P(n), i.e., P (n) : 1 + $\frac{1}{1+2}+\frac{1}{1+2+3}$ + ... + $rac{1}{1+2+3+\dots+n}$ = $rac{2n}{n+1}$ First we prove that the statement is true for n = 1. P (1) = $rac{2 imes1}{1+1}$ = $rac{2}{2}$ = 1 , which is true Assume that P(k) is true for some positive integer k, i.e., 1 + $\frac{1}{1+2}+\frac{1}{1+2+3}$ + ... + $rac{1}{1+2+3+\dots+k}$ = $rac{2k}{k+1}$ ... (i) We shall now prove that P(k + 1) is also true. For this we have to prove that 1 + $\frac{1}{1+2}+\frac{1}{1+2+3}$ + ... + $rac{1}{1+2+\dots+(k+1)}$ = $rac{2(k+1)}{(k+1)+1}$ L.H.S. = 1 + $\frac{1}{1+2}+\frac{1}{1+2+3}$ + ... + $rac{1}{1+2+\dots+(k+1)}$ = 1 + $\frac{1}{1+2}+\frac{1}{1+2+3}$ + ... + $rac{1}{1+2+\dots+k}$ + $rac{1}{1+2+\dots+(k+1)}$ = $rac{2k}{k+1}$ + $\frac{1}{\frac{(k+1)(k+2)}{2}}$ [From (i)] = $\frac{2k}{k+1} + \frac{2}{(k+1)(k+2)}$ = $rac{2k(k+2)+2}{(k+1)(k+2)}$ = $rac{2(k^2+2k+1)}{(k+1)(k+2)}$ = $rac{2(k+1)}{k+2}$ = R.H.S. Thus, P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.