Question

The sum of the following series to infinity $\frac{1}{\mathrm{1.3.5}}+\frac{1}{\mathrm{3.5.7}}+\frac{1}{\mathrm{5.7.9}}+...\infty is$ $\frac{1}{3c}$ .Find $c$

Answer 1

$\frac{1}{\mathrm{1.3.5}}+\frac{1}{\mathrm{3.5.7}}+\frac{1}{\mathrm{5.7.9}}+...\infty$
$=\frac{1}{4}(\frac{4}{\mathrm{1.3.5}}+\frac{4}{\mathrm{3.5.7}}+\frac{4}{\mathrm{5.7.9}}+...\infty )$
We can write,
$\frac{4}{\mathrm{1.3.5}}=\frac{1}{1.3}-\frac{1}{3.5}$
$\frac{4}{\mathrm{3.5.7}}=\frac{1}{3.5}-\frac{1}{5.7}$
Similarly, $\frac{4}{\mathrm{5.7.9}}=\frac{1}{5.7}-\frac{1}{7.9}$
So, the series can be written as
$=\frac{1}{4}(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+.....\infty )$
$=\frac{1}{4}\frac{1}{3}$
$=\frac{1}{12}$
Comparing with the given value ,
$c=4$