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Chapter 9: Problem 115

Find the indicated partial sum of the series. \(\sum_{n=1}^{\infty} 4\left(-\frac{1}{2}\right)^{n}\) Third partial sum

Short Answer

Expert verified
The third partial sum (S3) of the given geometric series is 4.66667 or approximately \( \frac{14}{3} \).

Step by step solution

01

Identify the series terms and common ratio

In this series, the first term 'a' is clearly 4 and the common ratio 'r' is \(-\frac{1}{2}\). So, we are trying to find the sum of the first three terms of the series 4, -2, and 1.
02

Use partial sum formula

To find the sum of the first three terms (n=3), we use the formula for the sum of the first 'n' terms of a geometric series, which is \( S_n = a (1 - r^n) / (1 - r) \). Substituting the identified values into the formula gives us \( S_3 = 4 * (1 - (-\frac{1}{2})^3) / {1 - (-\frac{1}{2})} \).
03

Solve for S3

Solving the equation gives us the third partial sum, S3. Perform the computations within the parentheses first, according to the order of operations (BIDMAS/BODMAS/PEDMAS rule), then proceed to the division. The result is the required third partial sum.

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Most popular questions from this chapter

$$1+\frac{1}{1}, 1+\frac{1}{2}, 1+\frac{1}{3}, 1+\frac{1}{4}, 1+\frac{1}{5}, \dots$$$$1+\frac{1}{1}, 1+\frac{1}{2}, 1+\frac{1}{3}, 1+\frac{1}{4}, 1+\frac{1}{5}, \dots$$

Write an expression for the apparent \(n\) th term of the sequence. (Assume \(n\) begins with \(1 .\)) $$9,11,13,15,17, \ldots$$

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