$1 + 3 + 7 + 15 + 31 + . . . . . . .$ to $n$ terms is equal to

2^{n + 1} - n - 2

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2^{n + 1} - n - 2 - 2

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2^{n} - n - 2

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None of these

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Solution

The correct option is A $2^{n + 1} - n - 2$
Let $T_{n}$ be the $n$ th term and $S$ be the sum up to $n$ terms
$S = 1 + 3 + 7 + 15 + 31 + . . . . . . + T_{n}$
Again, $S = 1 + 3 + 7 + 15 + . . . . . + T_{n - 1} + T_{n}$
On subtracting, we get
$0 = {1 + (2 + 4 + 8 + . . . . . . + (T_{n} - T_{n - 1})} - T_{n}$
$\Rightarrow T_{n} = 1 + 2 + 2^{2} + 2^{3} + . . . . . = \frac{1. (2^{n} - 1)}{2 - 1} = 2^{n} - 1$
Now $S = \sum T_{n} = \sum 2^{n} - \sum 1$
$= (2 + 2^{2} + 2^{3} + . . . . . + 2^{n}) - n$
$= 2 (\frac{2^{n} - 1}{2 - 1}) - n$
$= 2^{n + 1} - 2 - n$

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