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Chapter 8: Problem 11

Write the first four terms of each sequence whose general term is given. $$a_{n}=\frac{(-1)^{n+1}}{2^{n}-1}$$

Short Answer

Expert verified
The first four terms of the sequence are -1, 1/3, -1/7, 1/15.

Step by step solution

01

Understanding the sequence

The given general term of the sequence is \(a_{n}=\frac{(-1)^{n+1}}{2^{n}-1}\). This means that for each term of the sequence, we substitute the term number into 'n' in the general term to get the value of the term.
02

Getting the first term

To get the first term (n=1), we substitute n=1 into the general term to have \(a_{1} = \frac{(-1)^{1+1}}{2^{1}-1} = \frac{-1}{1}=-1\)
03

Getting the second term

Similarly, we can get the second term by substituting n=2 into the general term: \(a_{2} = \frac{(-1)^{2+1}}{2^{2}-1} = \frac{1}{3}\)
04

Getting the third term

Substituting n=3 into the general term gives us: \(a_{3} = \frac{(-1)^{3+1}}{2^{3}-1} = \frac{-1}{7}\)
05

Getting the fourth term

Substitute n=4 into the general term to get: \(a_{4} = \frac{(-1)^{4+1}}{2^{4}-1} = \frac{1}{15}\)

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