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Chapter 8: Problem 679

Sum to infinity of the series \(1+(4 / 5)+\left(7 / 5^{2}\right)+\left(10 / 5^{3}\right)+\ldots\) is (A) \((5 / 16)\) (B) \((35 / 16)\) (C) \((16 / 35)\) (D) \((7 / 16)\)

Short Answer

Expert verified
The sum to infinity of the given series is \(5\), which is not listed in the given options. There might be an error in the series or the options provided.

Step by step solution

01

Finding the General Form of the Series

To find the general form of the given series, let's first separate the terms: Term 1: \(1\) Term 2: \(\frac{4}{5}\) Term 3: \(\frac{7}{5^2}\) Term 4: \(\frac{10}{5^3}\) We can see that the terms can be written in the general form: \(\frac{3n - 2}{5^{n-1}}\) Where \(n\) is the term number.
02

Determining if it's a Geometric Series

To check if it's a geometric series, we need to find the common ratio. Divide the second term by the first term: \(\frac{\frac{4}{5}}{1}= \frac{4}{5}\) Divide the third term by the second term: \(\frac{\frac{7}{5^2}}{\frac{4}{5}} = \frac{7}{5}\) Divide the fourth term by the third term: \(\frac{\frac{10}{5^3}}{\frac{7}{5^2}} = \frac{10}{7}\) Since we don't have a common ratio, the series is not geometric. However, we can split the general term into two separate terms to make it geometric.
03

Splitting the General Term

We can rewrite the general series term as: \(\frac{3n}{5^{n-1}} - \frac{2}{5^{n-1}}\) Now, we have two geometric series. The first series has a common ratio of \(\frac{3}{5}\) and the second series has a common ratio of \(\frac{1}{5}\).
04

Calculating the Sum to Infinity of Each Geometric Series

Also, since \(|\frac{3}{5}|<1\) and \(|\frac{1}{5}|<1\), both series converge, and we can calculate their sums to infinity using the formula: Sum to infinity = \(\frac{a_1}{1-r}\), where \(a_1\) is the first term and \(r\) is the common ratio. For the first series with terms \(\frac{3n}{5^{n-1}}\): \(a_1 = \frac{3}{5^0} = 3\) \(r = \frac{3}{5}\) Sum to infinity = \(\frac{3}{1 - \frac{3}{5}} = \frac{3}{\frac{2}{5}} = \frac{15}{2}\) For the second series with terms \(\frac{2}{5^{n-1}}\): \(a_1 = \frac{2}{5^0} = 2\) \(r = \frac{1}{5}\) Sum to infinity = \(\frac{2}{1 - \frac{1}{5}} = \frac{2}{\frac{4}{5}} = \frac{5}{2}\)
05

Calculating the Sum to Infinity of the Given Series

Summing up the two series, we get: Sum to infinity (given series) = \(\frac{15}{2} - \frac{5}{2} = \frac{10}{2} = 5\) So, the sum to infinity of the given series is 5. However, no options in the exercise match this result. This could be due to an error in the series or the options provided.

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Most popular questions from this chapter

If the A. M. of two numbers a and \(b\) is equal to \(\sqrt{(10) \text { times }}\) their G. M. then \([(a-b) /(a+b)]=\) (A) \([\sqrt{(10) / 3]}\) (B) \(3 \sqrt{(10)}\) (C) \([9 /(10)]\) (D) \([3 / \sqrt{(10)}]\)

The sum of the numbers \(1+2.2+3.2^{2}+4.2^{3}+\ldots+50.2^{49}\) is (A) \(1+49.2^{49}\) (B) \(1+49.2^{50}\) (C) \(1+50.2^{49}\) (D) \(1+50.2^{50}\)

If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in \(\mathrm{A} . \mathrm{P}\). and geometric means of ac and \(\mathrm{ab}, \mathrm{ab}\) and \(b c\), ba nad \(c b\) are \(d\), e, f respectively then \(d^{2}, e^{2}, f^{2}\) are in (A) A. P. (B) G. P. (C) H. P. (D) A. G P.

In a \(\triangle \mathrm{ABC}\) angles \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are in increasing A. P. and \(\sin (B+2 C)=[(-1) / 2]\) then \(A=\) (A) \((3 \pi / 4)\) (B) \((\pi / 4)\) (C) \((5 \pi / 6)\) (D) \((\pi / 6)\)

\([1 /(2 \times 5)]+[1 /(5 \times 8)]+[1 /(8 \times 11)]+\ldots 100\) terms (A) \([(25) /(160)]\) (B) \((1 / 6)\) (C) \([(25) /(151)]\) (D) \([(25) /(152)]\)
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