Question

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

Answer 1

Let $P\left(n\right)(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})\cdots (1+\frac{(2n+1)}{n^2})=(n+1{)}^2$
For n = 1
$P\left(1\right)=1+\frac{(2\times 1+1)}{1^2}=(1+1{)}^2\Rightarrow 1+3=(2{)}^2\Rightarrow 4=4\therefore$ P(1) is true
Let P(n) be true for n = k
$\therefore$ P(k)
$(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})\cdots (1+\frac{(2k+1)}{k^2})=(k+1{)}^2\therefore \text{R.H.S.}=(k+2{)}^2\text{L.H.S}=(k+1{)}^2[1+\frac{(2k+3)}{(k+1{)}^2}]=(k+1{)}^2\left[\frac{(k+1{)}^2+(2k+3)}{(k+1{)}^2}\right]=k^2+4k+4=(k+2{)}^2$
$\therefore P(k+1)$ is true
thus P (k ) is true $\Rightarrow P(k+1)$ is true
Hence by principle fo mathematical induction,