$1 + 3 + 7 + 15 + 31 + . . . . . . . . . +$ to n terms=.

2^{n + 1} n

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2^{n + 1} n - 2

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2^{n} n - 2

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Solution

The correct option is B $2^{n + 1} n - 2$
Let $T_{n}$ is the nth term
Then sum upto n term is
$S = 1 + 3 + 7 + 15 + 31 + . . . . . . + T_{n}$ ...(1)
Again $S = 1 + 3 + 7 + 15 + . . . . . . + T_{n - 1} + T_{n}$ ...(2)
Subtracting (2) -(1) we get
$0 = 1 + [2 + 4 + 8 + 16 + . . . . . (T_{n} - T_{n 1})] - T_{n}$
$T_{n} = 1 + 2 + 2^{2} + 2^{3} + 2^{4} +$ upto n terms
Therefore
$S = \sum T_{n} = \sum 2^{n - 1} = (2 + 2^{2} + 2^{3} + . . . . . . + 2^{n}) n = 2 (\frac{2^{n} - 1}{2 - 1}) n = 2^{n + 1} n - 2 n$

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