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Chapter 8: Problem 35

Find the indicated sum. Use the formula for the sum of the first \(n\) terms of a geometric sequence. $$\sum_{i=1}^{6}\left(\frac{1}{2}\right)^{i+1}$$

Short Answer

Expert verified
The sum of the sequence is \(\frac{63}{32}\).

Step by step solution

01

Identifying the Parameters of the Sequence

The first term, \(a\), is obtained when \(i = 1\), thus \(a = (\frac{1}{2})^{1+1} = \frac{1}{4}\). The common ratio, \(r\), is the base which is \(\frac{1}{2}\). The number of terms, \(n\), is given as 6.
02

Applying the Sum Formula

We substitute \(a = \frac{1}{4}\), \(r = \frac{1}{2}\), and \(n = 6\) into the sum formula: \(S_n = \frac{a*(1 - r^n)}{1 - r} = \frac{\frac{1}{4}*(1 - (\frac{1}{2})^6)}{1 - \frac{1}{2}}\)
03

Solving the Expression

We calculate the expression to obtain the sum: \(S_n = \frac{\frac{1}{4}*(1 - \frac{1}{64})}{\frac{1}{2}} = \frac{\frac{63}{64}}{\frac{1}{2}} = \frac{63}{32}.\)

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