/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Evaluate the geometric series. ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 23

Evaluate the geometric series. $$ \frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots+\frac{1}{4^{50}} $$

Short Answer

Expert verified
The sum of the geometric series \(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots+\frac{1}{4^{50}}\) can be computed using the formula \(\text{Sum} = \frac{a(r^n - 1)}{r - 1}\) with \(a = \frac{1}{4}\), \(r = \frac{1}{4}\), and \(n = 51\). The sum is approximately 0.3333332.

Step by step solution

01

1. Identify the geometric series

We are given a series: \[ \frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots+\frac{1}{4^{50}} \] It's a geometric series because the terms can be expressed in the form: \(ar^n\), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the exponent, ranging from 0 to some positive integer.
02

2. Identify the first term and the common ratio

In this case, the first term, \(a\), is \(\frac{1}{4}\), and the common ratio, \(r\), can be found by dividing the second term by the first term: \[ r = \frac{\frac{1}{16}}{\frac{1}{4}} = \frac{1}{4} \]
03

3. Determine the number of terms in the series

To find the number of terms in the series, we need to identify the last term's exponent. The last term can be expressed as \[ \frac{1}{4^{50}} = a \cdot r^{n-1} \] Where \(n\) is the number of terms. Since \(a = \frac{1}{4}\) and \(r = \frac{1}{4}\), we get: \[ \frac{1}{4^{50}} = \left(\frac{1}{4}\right) \cdot \left(\frac{1}{4}\right)^{n-1} \] We can see that the exponent in the last term is 50: \[ n - 1 = 50 \Rightarrow n = 51 \] So, the series has 51 terms.
04

4. Apply the formula for the sum of a geometric series

The sum of a geometric series can be computed using the formula: \[ \text{Sum} = \frac{a(r^n - 1)}{r - 1} \] By plugging in our values \(a = \frac{1}{4}\), \(r = \frac{1}{4}\), and \(n = 51\), we get: \[ \text{Sum} = \frac{\frac{1}{4}\left(\left(\frac{1}{4}\right)^{51} - 1\right)}{\frac{1}{4} - 1} = -\frac{1}{3}\left(\frac{1}{4^{51}} - 1\right) \]
05

5. Calculate the sum

Finally, we can compute the sum: \[ \text{Sum} = -\frac{1}{3}\left(\frac{1}{4^{51}} - 1\right) \approx 0.3333332 \] Thus, the sum of the given geometric series is approximately 0.3333332.

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