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Chapter 19: Problem 5

Write out explicitly the series $$ \sum_{k=1}^{4} \frac{1}{(2 k+1)(2 k+3)} $$

Short Answer

Expert verified
$ Answer: The explicit representation of the series is $\frac{1}{15} + \frac{1}{35} + \frac{1}{63} + \frac{1}{99}$.

Step by step solution

01

Substitute values of k

First, substitute the values of k from 1 to 4 into the formula given: \(k=1: \frac{1}{(2 (1)+1)(2 (1)+3)}\) \(k=2: \frac{1}{(2 (2)+1)(2 (2)+3)}\) \(k=3: \frac{1}{(2 (3)+1)(2 (3)+3)}\) \(k=4: \frac{1}{(2 (4)+1)(2 (4)+3)}\)
02

Evaluate each term

Now, evaluate each term: \(k=1: \frac{1}{(2+1)(2+3)} = \frac{1}{3 \times 5}\) \(k=2: \frac{1}{(4+1)(4+3)} = \frac{1}{5 \times 7}\) \(k=3: \frac{1}{(6+1)(6+3)} = \frac{1}{7 \times 9}\) \(k=4: \frac{1}{(8+1)(8+3)} = \frac{1}{9 \times 11}\)
03

Add the terms

Now that we have evaluated each term, we can add them to find the sum: $$\sum_{k=1}^{4} \frac{1}{(2 k+1)(2 k+3)} = \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} + \frac{1}{9 \times 11}$$ Now the series is written out explicitly as: $$\frac{1}{15} + \frac{1}{35} + \frac{1}{63} + \frac{1}{99}$$

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Most popular questions from this chapter

An arithmetic sequence is given by \(b, \frac{2 b}{3}, \frac{b}{3}, 0, \ldots\) (a) State the sixth term. (b) State the \(k\) th term. (c) If the 20 th term has a value of 15 , find \(b\).

Find the seventh term of a geometric sequence with first term 2 and common ratio 3 .

A geometric sequence is given by \(1, \frac{1}{2}, \frac{1}{4}, \ldots\) What is its common ratio?

Find the sum of the squares of the first 20 positive integers.

A geometric sequence has first term 1 . The ninth term exceeds the fifth term by 240 . Find possible values for the eighth term.
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