Question

If S_n = \[\sum^n_{r = 1} \frac{1 + 2 + 2^2 + . . . \text { Sum to r terms }}{2^r}\], then S_n is equal to

Options

• 2ⁿ − n − 1

•   \[1 - \frac{1}{2^n}\] 

• \[n - 1 + \frac{1}{2^n}\]

• 2ⁿ − 1

Answer 1

\[n - 1 + \frac{1}{2^n}\]

We have:
S_n 

\[\sum^n_{r = 1} \frac{1 + 2 + 2^2 + . . . \text { sum to r terms }}{2^r}\]

\[ \Rightarrow S_n = \sum^n_{r = 1} \frac{1\left( 2^r - 1 \right)}{2^r}\]

\[ \Rightarrow S_n = \sum^n_{r = 1} \left( 1 - \frac{1}{2^r} \right)\]

\[ \Rightarrow S_n = n - \sum^n_{r = 1} \left( \frac{1}{2^r} \right)\]

\[ \Rightarrow S_n = n - \left[ \frac{\left( \frac{1}{2} \right)\left\{ 1 - \left( \frac{1}{2} \right)^n \right\}}{1 - \frac{1}{2}} \right]\]

\[ \Rightarrow S_n = n - \left[ 1 - \left( \frac{1}{2} \right)^n \right]\]

\[ \Rightarrow S_n = n - 1 + \frac{1}{2^n}\]