The sum of first $8$ terms of the series, whose $r^{t h}$ terms is $(2 r + 1) 2^{r}$ , is

2 (2^{12} - 2^{8} + 1)

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(2^{12} - 2^{8} + 1)

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(2^{12} - 2^{8} - 1)

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2 (2^{12} - 2^{8} - 1)

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Solution

The correct option is A $2 (2^{12} - 2^{8} + 1)$
$r^{t h} t e r m = (2 r + 1) 2^{r} S_{8} = 3.2 + 5. 2^{2} + 7.. 2^{3} + . . . . + 17. 2^{8} 2 S_{8} = 3. 2^{2} + 5. 2^{3} + . . . . + 15. 2^{8} + 17. 2^{9} S_{8} - 2 S_{8} = 3.2 + 2. 2^{2} + 2. 2^{3} + . . . . + 2. 2^{8} - 17. 2^{9} - S_{8} = 2 + 2^{2} + 2^{3} + 2^{4} + . . . . + 2^{9} - 17. 2^{9} - S_{8} = 2^{10} - 2 - 2^{13} - 2^{9} S_{8} = 2^{13} - 2^{9} + 2 S_{8} = 2 (2^{12} - 2^{8} + 1)$

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