The value of $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\cdots \left(1+\frac{2\mathrm{n}+1}{{\mathrm{n}}^2}\right)$ is 

detailed solution

Correct option is B

Let the given statement be P(n). i.eP(n):1+311+541+79…1+2n+1n2=(n+1)2Step l :For n=1,i.e.,P(1)=(1+1)2=22=4=1+31 which is true. Step lI: Let it is true for n = k, 1+311+541+79…1+2k+1k2=(k+1)2...(i) Step III For n= k + 1, 1+311+541+79…1+2k+1k21+2k+2+1(k+1)2=(k+1)21+2k+3(k+1)2  [using Eq. (i)] =(k+1)2(k+1)2+2k+3(k+1)2=k2+2k+1+2k+3=(k+2)2=[(k+1)+1]2∵(a+b)2=a2+2ab+b2Therefore, P(k + 1) is true when P(e) is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.