Question

$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....$ to $n$ terms $=\frac{n}{2n+1}$.

Answer 1

Now,

$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....$ to $n$ terms

$=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{(2n-1)(2n+1)}$

$=\frac{1}{2}([1-\frac{1}{3}]+[\frac{1}{3}-\frac{1}{5}]+[\frac{1}{5}-\frac{1}{7}]+....+[\frac{1}{(2n-1)}-\frac{1}{(2n+1)}])$

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

$=\frac{n}{2n+1}$.