$2 + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} + 2^{7} + 2^{8} + 2^{9} + 2^{10} + 2^{11}$ =___

$2^{11} - 2$

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$2^{12} - 2$

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$2^{11} - 1$

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$2^{12} - 1$

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Solution

To find:
$2 + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} + 2^{7} + 2^{8} + 2^{9} + 2^{10} + 2^{11}$
Here,
$\frac{t_{2}}{t_{1}} = \frac{2^{2}}{2} = 2$
$\frac{t_{3}}{t_{2}} = \frac{2^{3}}{2^{2}} = 2$
and clearly
$\frac{t_{2}}{t_{1}} = \frac{t_{3}}{t_{2}} = \frac{t_{4}}{t_{3}} = . . . .$
So, the given series is in Geometric progression.
Now, in $2 + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} + 2^{7} + 2^{8} + 2^{9} + 2^{10} + 2^{11}$
First term $(a) = 2$
Common ration $(r) = \frac{t_{2}}{t_{1}} = \frac{2^{2}}{2} = 2$
Number of terms $(n) = 11$
Since, Sum of the first $n$ -terms in G.P is,
$S_{n} = \frac{a (r^{n} - 1)}{r - 1}, r > 1$
$\Rightarrow S_{12} = \frac{2 (2^{11} - 1)}{2 - 1}$
$\Rightarrow S_{12} = {2.2}^{11} - 2$
$\Rightarrow S_{12} = 2^{12} - 2$
Hence, Option B is correct.

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2+22+23+24+25+26+27+28+29+210+211 =___

$2 + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} + 2^{7} + 2^{8} + 2^{9} + 2^{10} + 2^{11}$ =___