Question

If $(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})...(1+\frac{(2n+1)}{n^2})={a(n+1)}^2$.

Answer 1

Here $a_n=1+\frac{2n+1}{n^2}$

Now, we have $\prod _{n=1}^n{a}_n=a{\left(n+1\right)}^2$

$\Rightarrow \prod _{n=1}^n\left(1+\frac{2n+1}{n^2}\right)=a{\left(n+1\right)}^2$

$\Rightarrow \prod _{n=1}^n\left(\frac{n^2+2n+1}{n^2}\right)=a{\left(n+1\right)}^2$

$\Rightarrow \prod _{n=1}^n{\left(\frac{n+1}{n}\right)}^2=a{\left(n+1\right)}^2$

$\Rightarrow {\left(\prod _{n=1}^n\frac{n+1}{n}\right)}^2=a{\left(n+1\right)}^2$

$\Rightarrow {\left(\frac{2}{1}.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{n+1}{n}\right)}^2=a{\left(n+1\right)}^2$

$\Rightarrow {\left(n+1\right)}^2=a{\left(n+1\right)}^2$

$\Rightarrow a=1$