# Find the sum :

Question:

Find the sum : $\sum_{n=1}^{10}\left[\left(\frac{1}{2}\right)^{n-1}+\left(\frac{1}{5}\right)^{n+1}\right]$.

Solution:

$\sum_{n=1}^{10}\left[\left(\frac{1}{2}\right)^{n-1}+\left(\frac{1}{5}\right)^{n+1}\right]$

$=\sum_{n=1}^{10}\left(\frac{1}{2}\right)^{n-1}+\sum_{n=1}^{10}\left(\frac{1}{5}\right)^{n+1}$

$=\left\{1+\frac{1}{2}+\frac{1}{4}+\ldots+\left(\frac{1}{2}\right)^{9}\right\}+\left\{\frac{1}{5^{2}}+\frac{1}{5^{3}}+\frac{1}{5^{4}}+\ldots+\frac{1}{5^{11}}\right\}$

$=1\left(\frac{1-\left(\frac{1}{2}\right)^{10}}{1-\frac{1}{2}}\right)+\frac{1}{25}\left(\frac{1-\left(\frac{1}{5}\right)^{10}}{1-\frac{1}{5}}\right)$

$=\left(\frac{2^{10}-1}{2^{9}}\right)+\left(\frac{5^{10}-1}{4 \times 5^{11}}\right)$