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Chapter 7: Problem 24

Evaluate the geometric series. $$ \frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots+\frac{1}{3^{33}} $$

Short Answer

Expert verified
The sum of the given geometric series is equal to: \[ S_{33} = \frac{1}{2}(1 - \frac{1}{3^{33}}) \]

Step by step solution

01

Identify the geometric series

First, identify the given series as a geometric series. The series can be written as: \[ \frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{33}} \] Each term in the series is a power of 1/3, multiplied with the series of natural numbers in decreasing order. Therefore, this is a geometric series with a common ratio of 1/3.
02

Find the sum of the geometric series

To find the sum of a geometric series, we can use the formula: \[ S_n = \frac{a_{1}(1-r^n)}{1-r} \] where: - \(S_n\) = sum of the series - \(a_1\) = first term of the series - \(r\) = common ratio - \(n\) = number of terms in the series Here, \(a_1 = \frac{1}{3}\), \(r = \frac{1}{3}\), and \(n = 33\). Plug in the values and solve for the sum of the series.
03

Calculate the sum

Substitute the values into the formula and find the sum: \[ S_{33} = \frac{\frac{1}{3}(1 - (\frac{1}{3})^{33})}{1 - \frac{1}{3}} \] Simplify and solve for the sum: \[ S_{33} = \frac{\frac{1}{3}(1 - \frac{1}{3^{33}})}{\frac{2}{3}} \] \[ S_{33} = \frac{1}{2}(1 - \frac{1}{3^{33}}) \] Therefore, the sum of the given geometric series is equal to: \[ S_{33} = \frac{1}{2}(1 - \frac{1}{3^{33}}) \]

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