Find the sum of the series $4 + 4^{2} + 4^{3} + . . . {.4}^{100} + {4.4}^{101}$

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Solution

The correct option is C

Given series is
$s = 4 + 4^{2} + 4^{3} + . . . . {.4}^{100} + 4 X 4^{101}$
Up to 100 terms of series form a GP and $101^{t h}$ term is $4 \times 4^{101}$
$S = \frac{4 (4^{100} - 1)}{4 - 1} + 4 \times 4^{101}$
$= \frac{4^{100} - 4}{3} + 4 \times 4^{101}$
$= \frac{4^{100}}{3} - \frac{4}{3} + 4 \times 4^{101} = \frac{13 \times 4^{101}}{3} = \frac{{13.4}^{101} - 4}{3}$

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