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Chapter 8: Problem 9

Evaluate each geometric sum. $$\sum_{k=0}^{20}\left(\frac{2}{5}\right)^{2 k}$$

Short Answer

Expert verified
Question: Evaluate the geometric sum \(\displaystyle\sum_{k=0}^{20}\left(\frac{2}{5}\right)^{2 k}\). Answer: The sum of the given geometric series is approximately \(8.943\).

Step by step solution

01

Identify the sequence and common ratio

The given geometric sum is \(\displaystyle\sum_{k=0}^{20}\left(\frac{2}{5}\right)^{2 k}\). We notice that the common ratio of the series is \(r = \left(\frac{2}{5}\right)^2\) as the exponent of the term is \(2k\).
02

Use the geometric series sum formula

The formula for the sum of a geometric series is given by \(S_n = \frac{a_1(1-r^n)}{1-r}\), where \(n\) is the number of terms, \(a_1\) is the first term, and \(r\) is the common ratio.
03

Substitute the values into the sum formula

In our case, the first term \((a_1)\) is obtained when \(k = 0\), so \(a_1 = \left(\frac{2}{5}\right)^{0} = 1\). There are \(n= 21\) terms in the sequence as the series goes from \(k=0\) to \(k=20\), so we will substitute these values and the common ratio into the sum formula: \(S_{21} = \frac{1(1-\left(\frac{2}{5}\right)^{2 \cdot 20})}{1-\left(\frac{2}{5}\right)^2}\).
04

Simplify the expression

Now, we will simplify the expression to get the sum of the series: \(S_{21} = \frac{1(1-\left(\frac{2}{5}\right)^{40})}{1-\left(\frac{2}{5}\right)^2}\).
05

Evaluate the sum

After calculating the sum, we get: \(S_{21} = \frac{1(1-\left(\frac{2}{5}\right)^{40})}{\frac{21}{25}}\). Then by dividing each term in the numerator by the denominator, we obtain: \(S_{21} = \frac{1 - \left(\frac{2}{5}\right)^{40}}{\frac{21}{25}} = \frac{25}{21}(1- \left(\frac{2}{5}\right)^{40}) \approx 8.943\). So, the sum of the given geometric series is approximately \(8.943\).

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Most popular questions from this chapter

The famous Fibonacci sequence was proposed by Leonardo Pisano, also known as Fibonacci, in about \(\mathrm{A.D.} 1200\) as a model for the growth of rabbit populations. It is given by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1 .\) Each term of the sequence is the sum of its two predecessors. a. Write out the first ten terms of the sequence. b. Is the sequence bounded? c. Estimate or determine \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}},\) the ratio of the successive terms of the sequence. Provide evidence that \(\varphi=(1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Use induction to verify the remarkable result that $$f_{n}=\frac{1}{\sqrt{5}}\left(\varphi^{n}-(-1)^{n} \varphi^{-n}\right).$$

Consider the sequence \(\left\\{x_{n}\right\\}\) defined for \(n=1,2,3, \ldots\) by $$x_{n}=\sum_{k=n+1}^{2 n} \frac{1}{k}=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2 n}.$$ a. Write out the terms \(x_{1}, x_{2}, x_{3}\). b. Show that \(\frac{1}{2} \leq x_{n}<1,\) for \(n=1,2,3, \ldots\). c. Show that \(x_{n}\) is the right Riemann sum for \(\int_{1}^{2} \frac{d x}{x}\) using \(n\) subintervals. d. Conclude that \(\lim _{n \rightarrow \infty} x_{n}=\ln 2\).

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}} .\) When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots$$ Use the estimation techniques described in the text to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} 3^{-k}$$

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=\sqrt{n} \text { and } b_{n}=2 \ln n, n \geq 3$$
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