Question

Prove $(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\cdots (1+\frac{1}{n})=(n+1).$

Answer 1

$(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\cdots (1+\frac{1}{n})=(n+1).$
For n = 1
$P\left(1\right)=(1+\frac{1}{1})=1+1\Rightarrow 2=2\therefore P\left(1\right)\text{is true}\text{Let P (n) be true for n = k}\therefore P\left(k\right)=(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\cdots (1+\frac{1}{k})=(k+1).$
For n = k + 1
$R.H.S.=(k+1)(1+\frac{1}{k+1})=(k+1)\left[\frac{k+1+1}{k+1}\right]=(k+2)$
$\therefore$ P(k + 1 ) is true
Thus P (k) is true $\Rightarrow$ P(k + 1) is true
hence by principle of mathematical induction,