Chapter 15: Problem 5
Evaluate the following, giving each answer correct to two significant figures: (a) \(\sum_{n=1}^{20}(1.1)^{n}\) (b) \(\sum_{n=1}^{20} \log _{10}(1.1)^{n}\) (JMB)
Short Answer
Expert verified
(a) 62. (b) 8.7.
Step by step solution
01
- Identify the Series Type for Part (a)
Determine if the series given in Part (a), \(\sum_{n=1}^{20}(1.1)^{n}\), is geometric. The general form of a geometric series is \[ar^{0} + ar^{1} + ar^{2} + \ldots + ar^{n-1}\]. For this series, \a = 1.1\ and \r = 1.1.\
02
- Apply the Geometric Series Sum Formula for Part (a)
Use the sum formula for the first \[n\] terms of a geometric series, \[S_n = \frac{a(r^n - 1)}{r - 1}\]. Here, \a = 1.1, \r = 1.1, and \ = 20. Substitute these values into the formula: \[S_{20} = \frac{1.1(1.1^{20} - 1)}{1.1 - 1}.\]
03
- Calculate Part (a)
Calculate \1.1^{20}\ and then compute \S_{20}\. \[S_{20} = \frac{1.1((6.7275) - 1)}{0.1} \approx \frac{1.1(5.7275)}{0.1} \approx 62.9.\]
04
- Identify the Series Type for Part (b)
Determine if the series given in Part (b), \(\sum_{n=1}^{20} \log_{10}(1.1^n)\), can be simplified. Using the properties of logarithms: \log_{10}(1.1^n) = n \log_{10}(1.1).\
05
- Simplify the Series for Part (b)
Rewrite the sum in Part (b): \[ \sum_{n=1}^{20} n \log_{10}(1.1). \] Factoring out \log_{10}(1.1)\ gives: \ \log_{10}(1.1) \sum_{n=1}^{20} n.\
06
- Compute the Arithmetic Sum for Part (b)
Compute the summation \[ \sum_{n=1}^{20} n. \] The sum of the first \ positive integers is given by \[ \frac{n(n + 1)}{2},\] where \ = 20\. Substituting, get: \[ \frac{20(20 + 1)}{2} = 210.\]
07
- Final Calculation for Part (b)
Compute \ \log_{10}(1.1),\ which is approximately 0.0414. Then multiply by the sum: \[0.0414 \times 210 \approx 8.70.\]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
A geometric series is a sequence where each term is found by multiplying the previous term by a fixed, non-zero number called the common ratio. The general form is:
To calculate the sum of the first 'n' terms of a geometric series, the formula is: \[ S_n = \frac{a(r^n - 1)}{r - 1} \] This formula holds true as long as r ≠1. Apply the provided formula by plugging in the values for a, r, and n, and you will get the sum for part (a) of the exercise, \( \sum_{n=1}^{20}(1.1)^{n} \). Hence, we have a = 1.1, r = 1.1. The calculation is as follows: \[ S_{20} = \frac{1.1(1.1^{20} - 1)}{0.1} \] This simplifies to approximately 62.9.
- a, ar, ar^2, ar^3, \textellipsis
To calculate the sum of the first 'n' terms of a geometric series, the formula is: \[ S_n = \frac{a(r^n - 1)}{r - 1} \] This formula holds true as long as r ≠1. Apply the provided formula by plugging in the values for a, r, and n, and you will get the sum for part (a) of the exercise, \( \sum_{n=1}^{20}(1.1)^{n} \). Hence, we have a = 1.1, r = 1.1. The calculation is as follows: \[ S_{20} = \frac{1.1(1.1^{20} - 1)}{0.1} \] This simplifies to approximately 62.9.
Arithmetic Series
An arithmetic series is a sequence where the difference between consecutive terms is constant. This difference is called the ‘common difference’. Its general form is:
The sum of the first 'n' terms of an arithmetic series is given by the formula: \[ S_n = \frac{n}{2} (2a + (n - 1)d) \] However, in this exercise, for part (b), the terms are in an arithmetic progression in a slightly different way due to logarithmic properties. First, transform the series using: \( \log_{10}(1.1^n) = n \log_{10}(1.1) \). Then factor out \( \log_{10}(1.1) \) and sum the remaining arithmetic series: \[ \sum_{n=1}^{20} n \log_{10}(1.1) \] Compute the sum of the first 20 positive integers (an arithmetic series) using: \[ \sum_{n=1}^{20} n = \frac{20(20+1)}{2} = 210 \]. Hence, the series sum simplifies to: \( 0.0414 \times 210 = 8.70 \).
- a, a + d, a + 2d, a + 3d, \textellipsis
The sum of the first 'n' terms of an arithmetic series is given by the formula: \[ S_n = \frac{n}{2} (2a + (n - 1)d) \] However, in this exercise, for part (b), the terms are in an arithmetic progression in a slightly different way due to logarithmic properties. First, transform the series using: \( \log_{10}(1.1^n) = n \log_{10}(1.1) \). Then factor out \( \log_{10}(1.1) \) and sum the remaining arithmetic series: \[ \sum_{n=1}^{20} n \log_{10}(1.1) \] Compute the sum of the first 20 positive integers (an arithmetic series) using: \[ \sum_{n=1}^{20} n = \frac{20(20+1)}{2} = 210 \]. Hence, the series sum simplifies to: \( 0.0414 \times 210 = 8.70 \).
Logarithmic Properties
Logarithms are mathematical operations that are the inverse of exponentiation. They have several useful properties that help simplify complex expressions. Here are some key properties:
- \( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \) (Product Rule)
- \( \log_{b}(\frac{x}{y}) = \log_{b}(x) - \log_{b}(y) \) (Quotient Rule)
- \( \log_{b}(x^n) = n \log_{b}(x) \) (Power Rule)