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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: \[0.0414 \times 210 \approx 8.70.\]

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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:
  • a, ar, ar^2, ar^3, \textellipsis
where 'a' is the first term, and 'r' is the common ratio. For example, if you have a series 2, 4, 8, 16, \textellipsis, here 'a' is 2 and 'r' is 2.
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:
  • a, a + d, a + 2d, a + 3d, \textellipsis
where 'a' is the first term and 'd' is the common difference. For example, in the sequence 3, 6, 9, 12, \textellipsis, 'a' is 3 and 'd' is 3.
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)
In this exercise, the power rule is used for part (b), converting parts of the logarithmic series: \( \log_{10}(1.1^n) = n \log_{10}(1.1) \). This makes the computation easier, allowing you to factor out \( \log_{10}(1.1) \) and sum the series of integers. Knowing these properties not only helps in simplifying the problems but also in understanding the structure and behavior of logarithmic functions.

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Most popular questions from this chapter

The first term of an arithmetic series is \(\ln x\) and the \(r\) th term is \(\ln \left(x c^{r-1}\right)\) Show that the sum \(S_{n}\) of the first \(n\) terms of the series is \(\frac{n}{2} \ln \left(x^{2} c^{n-1}\right)\). \((\mathrm{AEB}) \mathrm{p}^{\prime} 76\)

The positive integers are bracketed as follows: $$ (1),(2,3),(4,5,6), \ldots $$ where there are \(r\) integers in the \(r\) th bracket. Find expressions for the first and last integers in the \(r\) th bracket. Find the sum of all the integers in the first 20 brackets. Prove that the sum of the integers in the \(r\) th bracket is \(\frac{1}{2}\left(r^{2}+1\right)\).

(a) Prove that \(\sum_{r=1}^{n} \frac{1}{r(r+1)}=\frac{n}{n+1}\) (b) Sum the series \(1+x+x^{2}+\ldots+x^{n}\) for \(x \neq 1\) By differentiation with respect to \(x\), or otherwise, find the value of $$ 1+2 x+3 x^{2}+\ldots+n x^{n-1} $$ and deduce the value of $$ 1.2+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots+n \cdot 2^{n} $$ (U of \(\mathrm{L})\)

(a) Assuming the formula for \(\sum_{r=1}^{n} r^{2}\), write down (i) the sum of the squares of the first \(2 n\) positive integers, (ii) the sum of the squares of the first \(n\) even integers. Hence find \(1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}\) (b) If \(S_{n}=\sum_{r=0}^{n} a^{r}\left(1+a+a^{2}+\ldots+a^{r}\right), \quad(|a| \neq 1)\) by considering \((1-a) S_{n}\), show that $$ S_{n}=\frac{1-a^{2 n+2}}{\left(1-a^{2}\right)(1-a)}-\frac{a^{n+1}\left(1+a^{n+1}\right)}{(1-a)^{2}} $$ State the set of values of \(a\) for which \(S_{n}\) approaches a limit as \(n \rightarrow \infty\) and find the sum to infinity of the series for these values of \(a\). (U of L)

Show that if \(f(r) \equiv r(r+1)(r+2)\) then $$ f(r)-f(r-1) \equiv 3 r(r+1) $$ Hence find the sum of the series $$ 2+6+12+\ldots+r(r+1)+\ldots+n(n+1) $$ \((\mathrm{U}\) of \(\mathrm{L}) \mathrm{p}\)

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