Question:
If $S=\frac{7}{5}+\frac{9}{5^{2}}+\frac{13}{5^{3}}+\frac{19}{5^{4}}+\ldots .$, then $160 \mathrm{~S}$ is equal to________.
Solution:
$\mathrm{S}=\frac{7}{5}+\frac{9}{5^{2}}+\frac{13}{5^{3}}+\frac{19}{5^{4}}+\ldots$
$\frac{1}{5} \mathrm{~S}=\frac{7}{5^{2}}+\frac{9}{5^{3}}+\frac{13}{5^{4}}+\ldots$
On subtracting
$\frac{4}{5} \mathrm{~S}=\frac{7}{5}+\frac{2}{5^{2}}+\frac{4}{5^{3}}+\frac{6}{5^{4}}+\ldots$
$\mathrm{S}=\frac{7}{4}+\frac{1}{10}\left(1+\frac{2}{5}+\frac{3}{5^{2}}+\ldots\right)$
$\mathrm{S}=\frac{7}{4}+\frac{1}{10}\left(1-\frac{1}{5}\right)^{-2}$
$=\frac{7}{4}+\frac{1}{10} \times \frac{25}{16}=\frac{61}{32}$
$\Rightarrow 160 S=5 \times 61=305$