Let the given statement be P(n), i.e., P (n) : $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ ... $\left(1+\frac{2n+1}{n^2}\right)$ = $(n+1)^2$ First we prove that the statement is true for n = 1. P (1) : $\left(1+\frac{3}{1}\right)$ = $(1+1)^2$ ⇒ 1 + 3 = $2^2$ ⇒ 4 = 4 , which is true Assume P(k) is true for some positive integer k, i.e., $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ ... $\left(1+\frac{2k+1}{k^2}\right)$ = $(k+1)^2$ ... (i) Now we shall prove that P(k + 1) is also true. For this we have to prove that $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ ... $\left(1+\frac{2k+1}{k^2}\right)$ $\left(1+\frac{2(k+1)+1}{(k+1)^2}\right)$ = $\left[(k+1)+1\right]^2$ L.H.S. = $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ ... $\left(1+\frac{2k+1}{k^2}\right)$ $\left(1+\frac{2(k+1)+1}{(k+1)^2}\right)$ = $(k+1)^2 \left(1+\frac{2(k+1)+1}{(k+1)^2}\right)$ [From (i)] = $\frac{(k+1)^2 \left[(k+1)^2+2(k+1)+1\right]}{(k+1)^2}$ = $(k+1)^2$ + 2 (k + 1) + 1 = $k^2$ + 2k + 1 = $k^2$ + 2k + 1 + 2k + 2 + 1 = $k^2$ + 4k + 4 = $(k+2)^2$ = $\left[(k+1)+1\right]^2$ = 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.