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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 10, Problema 10.3.9

9–15. Geometric sums Evaluate each geometric sum.


∑ k = 0 to 83ᵏ

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1
Identify the type of series given. Here, the sum is a geometric series because each term is obtained by multiplying the previous term by a constant ratio. The series is \( \sum_{k=0}^{8} 3^k \).
Recall the formula for the sum of the first \( n+1 \) terms of a geometric series: \[ S_{n} = \frac{a(r^{n+1} - 1)}{r - 1} \] where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the highest exponent index.
Identify the values for this problem: the first term \( a = 3^0 = 1 \), the common ratio \( r = 3 \), and the number of terms is \( n+1 = 9 \) since \( k \) goes from 0 to 8.
Substitute these values into the geometric sum formula: \[ S_8 = \frac{1 \times (3^{9} - 1)}{3 - 1} \]. This expression represents the sum of the series.
Simplify the denominator and prepare to evaluate the numerator (without calculating the final number), which completes the setup for finding the sum of the geometric series.

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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. For example, in the series 3^0 + 3^1 + ... + 3^8, the common ratio is 3.
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Geometric Series

Formula for the Sum of a Finite Geometric Series

The sum of the first n+1 terms of a geometric series with initial term a and common ratio r (r ≠ 1) is given by S = a(r^(n+1) - 1) / (r - 1). This formula allows quick calculation without adding each term individually.
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Geometric Series

Exponentiation and Index Notation

Exponentiation involves raising a base to a power, indicating repeated multiplication. Understanding how to interpret and manipulate expressions like 3^k is essential for evaluating terms in the geometric sum.
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