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Question
If Sn = \[\sum^n_{r = 1} \frac{1 + 2 + 2^2 + . . . \text { Sum to r terms }}{2^r}\], then Sn is equal to
Options
2n − n − 1
\[1 - \frac{1}{2^n}\]
\[n - 1 + \frac{1}{2^n}\]
2n − 1
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Solution
\[n - 1 + \frac{1}{2^n}\]
We have:
Sn
\[\sum^n_{r = 1} \frac{1 + 2 + 2^2 + . . . \text { sum to r terms }}{2^r}\]
\[ \Rightarrow S_n = \sum^n_{r = 1} \frac{1\left( 2^r - 1 \right)}{2^r}\]
\[ \Rightarrow S_n = \sum^n_{r = 1} \left( 1 - \frac{1}{2^r} \right)\]
\[ \Rightarrow S_n = n - \sum^n_{r = 1} \left( \frac{1}{2^r} \right)\]
\[ \Rightarrow S_n = n - \left[ \frac{\left( \frac{1}{2} \right)\left\{ 1 - \left( \frac{1}{2} \right)^n \right\}}{1 - \frac{1}{2}} \right]\]
\[ \Rightarrow S_n = n - \left[ 1 - \left( \frac{1}{2} \right)^n \right]\]
\[ \Rightarrow S_n = n - 1 + \frac{1}{2^n}\]
