First slide

Geometric progression

Question

The sum S_n to n terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

Moderate

A

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

$1 - 2^{- n}$
C

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

$2^{n} - 1$

Solution

We have
$\begin{array}{l} S_{n} = (1 - \frac{1}{2}) + (1 - \frac{1}{4}) + (1 - \frac{1}{2^{3}}) + \dots + (1 - \frac{1}{2^{n}}) \\ = n - (\frac{1}{2} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{n}}) \\ = n - \frac{\frac{1}{2} [1 - {(\frac{1}{2})}^{n}]}{1 - \frac{1}{2}} \\ = n - 1 + 2^{- n} \end{array}$

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