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Q.
The sum to $n$ terms of the series, $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots .$. is equal to
Sequences and Series
Solution:
$S =\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+\ldots . . \text { up to } n \text { terms }= n -\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{2^2}+\ldots . \text { up to } n \text { terms }\right)$