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Q.
Sum of $n$ terms of the series $\frac{2}{5}+\frac{22}{25}+\frac{122}{125}+\frac{622}{625}+\ldots \ldots .$. is
Sequences and Series
Solution:
Given sum $=\left(1-\frac{3}{5}\right)+\left(1-\frac{3}{25}\right)+\left(1-\frac{3}{125}\right)+\left(1-\frac{3}{625}\right)+\ldots \ldots$
$= n -\left(\frac{3}{5}+\frac{3}{5^2}+\frac{3}{5^3}+\ldots . . . n \text { terms }\right)= n -\frac{\frac{3}{5}\left(1-\frac{1}{5^{ n }}\right)}{1-\frac{1}{5}}= n -\frac{3}{4}\left(1-5^{- n }\right)= n -\frac{3}{4}+\frac{3}{4} 5^{- n }$