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Chapter 8: Problem 6

Write the first four terms of each sequence whose general term is given. $$a_{n}=\left(-\frac{1}{3}\right)^{n}$$

Short Answer

Expert verified
The first four terms of the sequence are: -1/3, 1/9, -1/27, and 1/81.

Step by step solution

01

Substitute n = 1 in the general term

Substitute n = 1 into the general term \(a_{n}=\left(-\frac{1}{3}\right)^{n}\) to get the first term of the sequence. This gives \(a_{1}=\left(-\frac{1}{3}\right)^{1} = -\frac{1}{3}.\)
02

Substitute n = 2 in the general term

Substitute n = 2 into the general term \(a_{n}=\left(-\frac{1}{3}\right)^{n}\) to get the second term of the sequence. This gives \(a_{2}=\left(-\frac{1}{3}\right)^{2} = \frac{1}{9}.\)
03

Substitute n = 3 in the general term

Substitute n = 3 into the general term \(a_{n}=\left(-\frac{1}{3}\right)^{n}\) to get the third term of the sequence. This gives \(a_{3}=\left(-\frac{1}{3}\right)^{3} = -\frac{1}{27}.\)
04

Substitute n = 4 in the general term

Substitute n = 4 into the general term \(a_{n}=\left(-\frac{1}{3}\right)^{n}\) to get the fourth term of the sequence. This gives \(a_{4}=\left(-\frac{1}{3}\right)^{4} = \frac{1}{81}.\)

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