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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 10, Problem 10.3.13

9–15. Geometric sums Evaluate each geometric sum.


{Use of Tech}∑ k = 0 to 9(−3/4)ᵏ

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1
Identify the type of series given. This is a geometric series where each term is of the form \(a r^k\), with \(a\) being the first term and \(r\) the common ratio.
Determine the first term \(a\) by substituting \(k=0\) into the term \(\left(-\frac{3}{4}\right)^k\). Since any number to the zero power is 1, \(a = 1\).
Identify the common ratio \(r\) as \(-\frac{3}{4}\), which is the base being raised to the power \(k\).
Use the formula for the sum of the first \(n+1\) terms of a geometric series: \(S_{n} = a \frac{1 - r^{n+1}}{1 - r}\) where \(n=9\) in this problem.
Substitute \(a = 1\), \(r = -\frac{3}{4}\), and \(n = 9\) into the formula to express the sum as \(S_9 = \frac{1 - \left(-\frac{3}{4}\right)^{10}}{1 - \left(-\frac{3}{4}\right)}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Geometric Series

A geometric series is the sum of the terms of a geometric sequence, where each term is found by multiplying the previous term by a constant ratio. The series has the form ∑ ar^k, where a is the first term and r is the common ratio.
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Sum Formula for Finite Geometric Series

The sum of the first n+1 terms of a geometric series is given by S_n = a(1 - r^(n+1)) / (1 - r), provided r ≠ 1. This formula allows quick calculation of the sum without adding each term individually.
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Use of Technology in Calculations

Technology such as calculators or software can efficiently compute sums of series, especially when dealing with complex ratios or many terms. It helps verify manual calculations and handle more complicated expressions.
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