The sum of first $n$ terms of the series $\frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots$ is
The sum of first $n$ terms of the series $\frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots$ is
- $n+\frac{2^{n+1}}{3 \times 5^n}-\frac{2}{3}$
- $n-\frac{2^{n+1}}{3 \times 5^n}-\frac{2}{3}$
- $n+\frac{2^{n+1}}{3 \times 5^n}+\frac{2}{3}$
- $n-\frac{2^{n+1}}{3 \times 5^n}+\frac{2}{3}$
Solution
Given, series
$
\begin{aligned}
& \frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots+\text { upto } n \text { terms } \\
& =\left(1-\frac{2}{5}\right)+\left(1-\frac{4}{25}\right)+\left(1-\frac{8}{125}\right)+\ldots+\text { upto }
\end{aligned}
$
$n$ terms
$
\begin{aligned}
& =n-\left[\frac{2}{5}+\left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right)^3+\ldots+\text { upto } n \text { terms }\right] \\
& =n-\frac{\frac{2}{5}\left(1-\left(\frac{2}{5}\right)^n\right)}{1-\frac{2}{5}} \\
& =n-\frac{\left[\because \because a+a r+a r^2+\ldots+a r^{n-1}=\frac{a\left(1-r^n\right)}{1-r}\right]}{3 / 5}=n+\frac{2^{n+1}}{3 \times 5^n}-\frac{2}{3}
\end{aligned}
$
Hence, option (a) is correct
Asked in: AP EAMCET 2019 (20 Apr Shift 2)
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