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Chapter 10: Problem 4

Write the first five terms of the sequence. $$ a_{n}=\left(-\frac{1}{2}\right)^{n} $$

Short Answer

Expert verified
The first five terms of the sequence defined by \(a_{n}=\left(-\frac{1}{2}\right)^{n}\) are -1/2, 1/4, -1/8, 1/16, -1/32.

Step by step solution

01

Compute first term

The first term \(a_{1}\) is computed by substituting \(n=1\) into the formula. So \(a_{1}=\left(-\frac{1}{2}\right)^{1} = -\frac{1}{2}\.
02

Compute second term

The second term \(a_{2}\) is computed by substituting \(n=2\) into the formula. So \(a_{2}=\left(-\frac{1}{2}\right)^{2} =\frac{1}{4}\.
03

Compute third term

The third term \(a_{3}\) is computed by substituting \(n=3\) into the formula. So \(a_{3}=\left(-\frac{1}{2}\right)^{3} =-\frac{1}{8}\.
04

Compute fourth term

The fourth term \(a_{4}\) is computed by substituting \(n=4\) into the formula. So \(a_{4}=\left(-\frac{1}{2}\right)^{4} = \frac{1}{16}\.
05

Compute fifth term

The fifth term \(a_{5}\) is computed by substituting \(n=5\) into the formula. So \(a_{5}=\left(-\frac{1}{2}\right)^{5} = -\frac{1}{32}\.

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Most popular questions from this chapter

Verify that the infinite series diverges. $$ \sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n}=1+\frac{4}{3}+\frac{16}{9}+\frac{64}{27}+\cdots $$

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