The sum of the series $(1 + 2) + (1 + 2 + 2^{2}) + (1 + 2 + 2^{2} + 2^{3}) + . . .$ upto n terms is

2^{n + 2} - n - 4

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2 (2^{n} - 1) - n

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2^{n + 1} - n

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2^{n + 1} - 1

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Solution

The correct option is A $2^{n + 2} - n - 4$
The given series is
$(1 + 2) + (1 + 2 + 2^{2}) + (1 + 2 + 2^{2} + 2^{3}) + . . . + n t h t e r m$
$∴ T_{n} = 1 + 2 + 2^{2} + 2^{3} + . . . + 2^{n}$
$= \frac{1 (2^{n + 1} - 1)}{2 - 1} = 2^{n + 1} - 1$
$\sum T_{n} = \sum 2^{n + 1} - 1 = \sum 2^{n + 1} - \sum 1$
$= (2^{2} + 2^{3} + . . . + 2^{n + 1}) - n$
$= 2 (2 + 2^{2} + . . . + 2^{n}) - n$
$= \frac{4 (2^{n} - 1)}{2 - 1} - n = 2^{n + 2} - n - 4$

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The sum of the series $(1 + 2) + (1 + 2 + 2^{2}) + (1 + 2 + 2^{2} + 2^{3}) + . . .$ upto n terms is