First slide

Special Series in Sequences and Series

Question

$Let S = \frac{4}{19} + \frac{44}{19^{2}} + \frac{444}{19^{3}} + \dots . upto \infty . Then S is equal to$

Moderate

A

$40 / 9$
B

38/81
C

36/171
D

81/38

Solution

$S = \frac{4}{19} + \frac{44}{19^{2}} + \frac{444}{19^{3}} + .....$
$\Rightarrow \frac{1}{19} S = \frac{4}{19^{2}} + \frac{44}{19^{3}} + .....$
$\begin{matrix} Substracting (2) from (1), we get \\ \frac{18}{19} S = \frac{4}{19} + \frac{40}{19^{2}} + \frac{400}{19^{3}} + \dots . \\ = \frac{\frac{4}{19}}{1 - \frac{10}{19}} \\ = 4 / 9 \\ \Rightarrow S = 38 / 81 \end{matrix}$

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