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Chapter 14: 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 \(1, -1/3, 1/7, -1/15\) respectively.

Step by step solution

01

Substitute n = 1 into the general term

Substitute \(n = 1\) into the general term \(a_{n}=\frac{(-1)^{n+1}}{2^{n}-1}\). After substituting, the equation becomes \(a_{1}=\frac{(-1)^{1+1}}{2^{1}-1}\), which simplifies to \(a_{1} = 1\).
02

Substitute n = 2 into the general term

Substitute \(n = 2\) into the general term \(a_{n}=\frac{(-1)^{n+1}}{2^{n}-1}\). After substituting, the equation becomes \(a_{2}=\frac{(-1)^{2+1}}{2^{2}-1}\), which simplifies to \(a_{2} = -1/3\).
03

Substitute n = 3 into the general term

Substitute \(n = 3\) into the general term \(a_{n}=\frac{(-1)^{n+1}}{2^{n}-1}\). After substituting, the equation becomes \(a_{3}=\frac{(-1)^{3+1}}{2^{3}-1}\), which simplifies to \(a_{3} = 1/7\).
04

Substitute n = 4 into the general term

Substitute \(n = 4\) into the general term \(a_{n}=\frac{(-1)^{n+1}}{2^{n}-1}\). After substituting, the equation becomes \(a_{4}=\frac{(-1)^{4+1}}{2^{4}-1}\), which simplifies to \(a_{4} = -1/15\).

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