Question

Prove using PMI
$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)...\left(1+\frac{\left(2n+1\right)}{n^2}\right)={\left(n+1\right)}^2$

Answer 1

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

for $n=1,1+\frac{3}{1}=\frac{4}{1}={\left(1+1\right)}^2$

Let the statement be true for $n=k$

$\Rightarrow \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)...\left(1+\frac{\left(2k+1\right)}{k^2}\right)={\left(k+1\right)}^2$

for $n=k+1$,

$\begin{array}{c}\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)...\left(1+\frac{\left(2k+1\right)}{k^2}\right)\left(1+\frac{\left(2k+3\right)}{{\left(k+1\right)}^2}\right)\\ ={\left(k+1\right)}^2\left(1+\frac{\left(2k+3\right)}{{\left(k+1\right)}^2}\right)\\ ={\left(k+1\right)}^2\left(\frac{k^2+2k+1+2k+3}{{\left(k+1\right)}^2}\right)\\ ={\left(k+2\right)}^2\end{array}$

$\therefore$ The statement is true by PMI