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Chapter 8: Problem 4

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 \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), \(\frac{1}{81}\).

Step by step solution

01

Understand the given formula

The term \(a_{n}=\left(\frac{1}{3}\right)^{n}\) represents the nth term in the sequence. This means that by substituting different values of n, different terms of the sequence can be found.
02

Find the first term of the sequence

For the first term in the sequence, substitute n = 1 into the formula. This will give: \(a_{1}=\left(\frac{1}{3}\right)^{1} = \frac{1}{3}\). So, the first term is \(\frac{1}{3}\).
03

Find the second term of the sequence

Replace n where n = 2 in the formula. This calculates as: \(a_{2}=\left(\frac{1}{3}\right)^{2} = \frac{1}{9}\). Therefore, the second term of the sequence is \(\frac{1}{9}\).
04

Find the third term of the sequence

Insert n = 3 into the formula. This calculates as: \(a_{3}=\left(\frac{1}{3}\right)^{3} = \frac{1}{27}\). Thus, the third term in the sequence is \(\frac{1}{27}\).
05

Find the fourth term of the sequence

Finally, incorporate n = 4 into the formula. This operates out to be: \(a_{4}=\left(\frac{1}{3}\right)^{4} = \frac{1}{81}\). So, \(\frac{1}{81}\) becomes the fourth term in the sequence.

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