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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 10.3.11

9–15. Geometric sums Evaluate each geometric sum.


{Use of Tech}∑ k = 0 to 20(2/5)²ᵏ

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Identify the type of series given. The sum is a geometric series because each term is obtained by multiplying the previous term by a constant ratio.
Write down the general form of a geometric series sum: \(S_n = a \frac{1 - r^{n+1}}{1 - r}\), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms minus one.
Determine the first term \(a\) by substituting \(k=0\) into the term expression: \(a = \left(\frac{2}{5}\right)^{2 \cdot 0} = \left(\frac{2}{5}\right)^0\).
Identify the common ratio \(r\) by examining the factor that each term is multiplied by to get the next term. Since the exponent increases by 2 each time, \(r = \left(\frac{2}{5}\right)^2\).
Substitute \(a\), \(r\), and \(n=20\) into the geometric sum formula and simplify the expression to find the sum without calculating the final numerical value.

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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. The series has the form a + ar + ar² + ... + arⁿ, where a is the first term and r is the common ratio.
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Geometric Series

Sum Formula for a Finite Geometric Series

The sum of the first n+1 terms of a geometric series is given by S = 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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Geometric Series

Evaluating Powers and Substitution

To evaluate the geometric sum, you must correctly compute powers of the common ratio and substitute values into the sum formula. Understanding exponentiation and careful substitution ensures accurate calculation of the sum.
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Substitution With an Extra Variable