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(B) Let $T_n$ be the $n^{th}$ term and $S_n$ be the sum up to $n$ terms.$S_n = 1 + 3 + 7 + 15 + 31 + \dots + T_n$We can write the $n^{th}$ term as $T_

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

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(B) Let $T_n$ be the $n^{th}$ term and $S_n$ be the sum up to $n$ terms.$S_n = 1 + 3 + 7 + 15 + 31 + \dots + T_n$We can write the $n^{th}$ term as $T_n = 2^n - 1$.Thus,$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (2^k - 1)$.$S_n = \sum_{k=1}^{n} 2^k - \sum_{k=1}^{n} 1$.Using the sum formula for a geometric progression,$\sum_{k=1}^{n} 2^k = 2(2^n - 1) / (2 - 1) = 2^{n+1} - 2$.Therefore,$S_n = (2^{n+1} - 2) - n = 2^{n+1} - n - 2$.

$1 + 3 + 7 + 15 + 31 + \dots$ to $n$ terms =

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