Question

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

• A. $n-1$
• B. $n$
• C. $n^2$
• D. $n+1$

Answer 1

The correct option is D $n+1$

Taking $n=1$
$1+\frac{1}{1}=2$
$n-1=0,n^2=1,n+1=2$
$P\left(n\right):(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\cdots (1+\frac{1}{n})=n+1$
$P\left(1\right)$ is true.
Let $P\left(k\right)$ be true.
$\Rightarrow (1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\cdots (1+\frac{1}{k})=k+1$
$P(k+1):(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\cdots (1+\frac{1}{k})(1+\frac{1}{k+1})$
$=(k+1)\left(\frac{k+1+1}{k+1}\right)$
$=k+2$
$=(k+1)+1$
$P(k+1)$ is true.
$\therefore P\left(n\right)$ is true $\forall n\in N$