Question

$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\dots$ what will be the $n^{th}$ term?

Answer 1

$\left(\frac{1}{1\cdot 2}\right)+\left(\frac{1}{2\cdot 3}\right)+........+\left(\frac{1}{n(n+1)}\right)=\left(\right(\frac{1}{1}-\left(\frac{1}{2}\right))+((\frac{1}{2}-(\frac{1}{3}\left)\right)+...........+(\frac{1}{n}-(\frac{1}{n+1})=1-(\frac{1}{n+1})=(\frac{n+1-1}{n+1})=(\frac{n}{n+1})$