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Chapter 13: Problem 62

Determine whether each infinite geometric series converges or diverges. If it converges, find its sum. $$ \sum_{k=1}^{\infty} 8\left(\frac{1}{3}\right)^{k-1} $$

Short Answer

Expert verified
\( S = 12 \)

Step by step solution

01

- Identify the first term and common ratio

In an infinite geometric series, the general form is given by \( a r^{k-1}\) , where \(a\) is the first term and \(r\) is the common ratio. Here, the first term \(a = 8\) and the common ratio \(r = \frac{1}{3}\).
02

- Determine Convergence or Divergence

An infinite geometric series converges if and only if the absolute value of the common ratio is less than 1, that is \(|r| < 1\). In this case, \(r = \frac{1}{3}\), and since \(\frac{1}{3} < 1\), the series converges.
03

- Use the Sum Formula for Convergent Series

When a geometric series converges, its sum can be found using the formula for the sum of an infinite geometric series, \(S = \frac{a}{1-r}\). Substituting the known values, \(a = 8\) and \(r = \frac{1}{3}\), the sum is \( S = \frac{8}{1 - \frac{1}{3}} \) .
04

- Simplify the Sum

Now, simplify the equation \( S = \frac{8}{1 - \frac{1}{3}}\). This simplifies to \( S = \frac{8}{\frac{2}{3}} \) , which further simplifies to \( S = 8 \times \frac{3}{2} = 12\). Therefore, the sum of the series is 12.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

infinite geometric series
An infinite geometric series is a type of series where you keep adding terms indefinitely. Each term in the series is obtained by multiplying the previous term by a constant factor known as the common ratio. The series can be written in the form:

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Most popular questions from this chapter

Bode's Law In \(1772,\) Johann Bode published the following formula for predicting the mean distances, in astronomical units (AU), of the planets from the sun: $$ a_{1}=0.4 \quad a_{n}=0.4+0.3 \cdot 2^{n-2} $$ where \(n \geq 2\) is the number of the planet from the sun. (a) Determine the first eight terms of the sequence. (b) At the time of Bode's publication, the known planets were Mercury \((0.39 \mathrm{AU}),\) Venus \((0.72 \mathrm{AU}),\) Earth \((1 \mathrm{AU})\) Mars \((1.52 \mathrm{AU}),\) Jupiter \((5.20 \mathrm{AU}),\) and Saturn \((9.54 \mathrm{AU})\) How do the actual distances compare to the terms of the sequence? (c) The planet Uranus was discovered in \(1781,\) and the asteroid Ceres was discovered in \(1801 .\) The mean orbital distances from the sun to Uranus and Ceres " are \(19.2 \mathrm{AU}\) and \(2.77 \mathrm{AU},\) respectively. How well do these values fit within the sequence? (d) Determine the ninth and tenth terms of Bode's sequence. (e) The planets Neptune and Pluto" were discovered in 1846 and \(1930,\) respectively. Their mean orbital distances from the sun are \(30.07 \mathrm{AU}\) and \(39.44 \mathrm{AU},\) respectively. How do these actual distances compare to the terms of the sequence? (f) On July \(29,2005,\) NASA announced the discovery of a dwarf planet \((n=11),\) which has been named Eris. Use Bode's Law to predict the mean orbital distance of Eris from the sun. Its actual mean distance is not yet known, but Eris is currently about 97 astronomical units from the sun.

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1 \cdot 2+3 \cdot 4+5 \cdot 6+\cdots+(2 n-1)(2 n)=\frac{1}{3} n(n+1)(4 n-1) $$

Challenge Problem Use the Principle of Mathematical Induction to prove that $$ \left[\begin{array}{rr} 5 & -8 \\ 2 & -3 \end{array}\right]^{n}=\left[\begin{array}{cr} 4 n+1 & -8 n \\ 2 n & 1-4 n \end{array}\right] $$ for all natural numbers \(n\).

Determine whether the given sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the common difference; if it is geometric, find the common ratio. If the sequence is arithmetic or geometric, find the sum of the first 50 terms. $$ 2,4,6,8, \ldots $$

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ -2-3-4-\cdots-(n+1)=-\frac{1}{2} n(n+3) $$
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