Evaluate:
$2 + 2^{2} + 2^{3} + . . . . + 2^{9} =$

1396

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1022

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1587

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1478

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Solution

The correct option is A $1022$
$2 + 2^{2} + 2^{3} + . . . . . . + 2^{9}$
The series is in $G P$ with common difference $= \frac{2^{2}}{2} = \frac{2^{3}}{2} . . . . . = \frac{2^{9}}{2^{8}} = 2$
Sum of $G P = \frac{a (r^{n} - 1)}{r - 1}$ where $a$ is the first term and $r$ is the common difference and last term $= a (r^{n} - 1)$
So,Last term,
$2. 2^{n - 1} = 2^{9} 2^{n - 1} = 2^{8} n - 1 = 8 n = 9$
$∴ S_{9} = \frac{2 (2^{9} - 1)}{2 - 1} = 2 (512 - 1) = 2 (511) = 1022$
Answer $(B)$ .

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