# Summation of $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots$ till $n$ terms

What is the pattern in the following?

• Sum to $n$ terms of the series: $$\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots$$
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Hint: The $k$th term is $1-\frac1{2^k}$. – Did Jun 3 '12 at 15:39

Hint:

Write it as $(1-{1\over2})+(1-{1\over4})+(1-{1\over 8})+\cdots+(1-{1\over 2^n}).$

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Here is the pattern:

\begin{align*} \frac{1}{2} + \frac{3}{4} + \cdots &= \biggl(1-\frac{1}{2}\biggr) + \biggl(1-\frac{1}{2^2}\biggr) + \cdots \\\ &= (1+1+\cdots + 1) - \biggl(\frac{1}{2}+\frac{1}{2^2} + \cdots +\frac{1}{2^n}\biggr) \end{align*}

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So, what you want is basically $$\left(1 - \frac {1} {2}\right) + \left(1 - \frac {1} {4}\right) + \cdots + \left(1 - \frac {1} {2^n}\right) = n - \left(\frac {1} {2} + \frac {1} {4} + \cdots + \frac {1} {2^n}\right) = n - 1 + \frac {1} {2^n}.$$

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