Answer
See proof
Work Step by Step
We have to prove the statement:
$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right).....\left(1+\dfrac{1}{n}\right)=n+1$
Step 1: Prove that the statement is true for $n=1$:
The left side is:
$1+\dfrac{1}{1}=2$
The right side is:
$1+1=2$
As $2=2$, the statement is true for $n=1$.
Step 2: Assume the statement is true for $k$ implies it is true for $k+1$ too.
We have to prove:
$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right).....\left(1+\dfrac{1}{k}\right)\left(1+\dfrac{1}{k+1}\right)=(k+1)+1$
$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right).....\left(1+\dfrac{1}{k}\right)\left(1+\dfrac{1}{k+1}\right)=k+2$
Because the statement is true for $k$ we have:
$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right).....\left(1+\dfrac{1}{k}\right)=k+1$
Multiply both sides by $\left(1+\dfrac{1}{k+1}\right)$:
$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right).....\left(1+\dfrac{1}{k}\right)\left(1+\dfrac{1}{k+1}\right)=(k+1)\left(1+\dfrac{1}{k+1}\right)$
$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right).....\left(1+\dfrac{1}{k}\right)\left(1+\dfrac{1}{k+1}\right)=(k+1)\cdot\dfrac{k+2}{k+1}$
$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right).....\left(1+\dfrac{1}{k}\right)\left(1+\dfrac{1}{k+1}\right)=k+2$
We got that the statement is also true for $k+1$.
So, the given statement is true.
