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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 12: Q 39. (page 825)

In Problems 39鈥� 44, find each sum.
$\frac{1}{4} + \frac{2}{4} + \frac{2^{2}}{4} + \frac{2^{3}}{4} + . . . + \frac{2^{14}}{4}$

Short Answer

Expert verified

The sum of the given sequence is $\frac{1}{4} 路 - (1 - 2^{n})$

Step by step solution

01

Step 1. Write the given information.

The given sequence is $\frac{1}{4} + \frac{2}{4} + \frac{2^{2}}{4} + \frac{2^{3}}{4} + . . . + \frac{2^{14}}{4}$

02

Step 2. Find the common ratio.

$a_{1} = \frac{1}{4}, a_{2} = \frac{3}{4}, a_{3} = \frac{2^{2}}{4} \frac{a_{2}}{a_{1}} = \frac{\frac{2}{4}}{\frac{1}{4}} = 2 \frac{a_{3}}{a_{2}} = \frac{\frac{2^{2}}{4}}{\frac{2}{4}} = 2 r = 2$

03

Step 3. Find the sum of nth term.

The formula for the sum of nth terms $s_{n} = \frac{a_{1} (1 - r^{n})}{1 - r}$

$s_{n} = \frac{1}{4} 路 \frac{1 - 2^{n}}{1 - 2} = \frac{1}{4} 路 - (1 - 2^{n})$

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