The sum of the first n terms of the series fr | vedclass

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Question

(A) The given series is $\frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots$ up to $n$ terms.We can rewrite the $k$-th term as $1 - (\frac{2}{5})^k$.Thu

Vedclass · Accepted Answer

$n+\frac{2^{n+1}}{3 	imes 5^n}-\frac{2}{3}$

Vedclass · Answer

(A) The given series is $\frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots$ up to $n$ terms.We can rewrite the $k$-th term as $1 - (\frac{2}{5})^k$.Thus,the sum $S_n = \sum_{k=1}^{n} (1 - (\frac{2}{5})^k) = \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} (\frac{2}{5})^k$.$S_n = n - [\frac{2}{5} + (\frac{2}{5})^2 + \ldots + (\frac{2}{5})^n]$.This is a geometric progression with first term $a = \frac{2}{5}$ and common ratio $r = \frac{2}{5}$.The sum of $n$ terms is $\frac{a(1-r^n)}{1-r} = \frac{\frac{2}{5}(1-(\frac{2}{5})^n)}{1-\frac{2}{5}} = \frac{\frac{2}{5}(1-(\frac{2}{5})^n)}{\frac{3}{5}} = \frac{2}{3}(1-\frac{2^n}{5^n}) = \frac{2}{3} - \frac{2^{n+1}}{3 	imes 5^n}$.Substituting this back into $S_n = n - [\frac{2}{3} - \frac{2^{n+1}}{3 	imes 5^n}]$,we get $S_n = n - \frac{2}{3} + \frac{2^{n+1}}{3 	imes 5^n}$.Therefore,the correct option is $A$.

The sum of the first $n$ terms of the series $\frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots$ is

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