Question

Find the sum to infinite terms of the series
$\frac{7}{5}(1+\frac{1}{{10}^2}+\frac{1.3}{1.2}\cdot \frac{1}{{10}^4}+\frac{\mathrm{1.3.5}}{\mathrm{1.2.3}}\cdot \frac{1}{{10}^6}+......)$

Answer 1

$\frac{7}{5}(1+{\left(\frac{1}{{10}^2}\right)}^1+\frac{1.3}{1.2}\times {\left(\frac{1}{{10}^2}\right)}^2+...)$

$\frac{7}{5}(1+\frac{\frac{1}{2}}{1!}\times 2^1.{\left(\frac{1}{{10}^2}\right)}^1+\frac{\frac{1}{2}.\frac{3}{2}}{2!}\times 2^2.{\left(\frac{1}{{10}^2}\right)}^2+\frac{\frac{1}{2}.\frac{3}{2}.\frac{5}{2}}{3!}\times 2^3.{\left(\frac{1}{{10}^2}\right)}^3...)$

$=\frac{7}{5}(1+\frac{\frac{1}{2}}{1!}\times {\left(\frac{2}{{10}^2}\right)}^1+\frac{\frac{1}{2}.(\frac{1}{2}+1)}{2!}\times {\left(\frac{2}{{10}^2}\right)}^2+\frac{\frac{1}{2}.(\frac{1}{2}+1).(\frac{1}{2}+2)}{3!}\times {\left(\frac{2}{{10}^2}\right)}^3...)$

$=\frac{7}{5}{(1-\frac{2}{{10}^2})}^{\frac{-1}{2}}$ $[\because {(1-x)}^{-n}=1+\frac{nx}{1!}+\frac{n(n+1)}{2!}{x}^2+...]$

$=\frac{7}{5}{\left(\frac{98}{100}\right)}^{\frac{-1}{2}}$

$=\frac{7}{5}\times \frac{10}{7\sqrt{2}}$

$=\sqrt{2}$