Chapter 12: Q. 32 (page 1173)

In the following exercise, find a general term for the sequence whose first five terms are shown.
$\frac{1}{1}, \frac{1}{8}, \frac{1}{27}, \frac{1}{64}, \frac{1}{125}, \dots$

Short Answer

Expert verified

The general term for the sequence whose first five terms are $\frac{1}{1}, \frac{1}{8}, \frac{1}{27}, \frac{1}{64}, \frac{1}{125}, \dots$ is $a_{n} = \frac{1}{n^{3}} .$

Step by step solution

Step 1. Given information

$\begin{array}{l} Thefirstfivetermsofthesequenceare \\ \frac{1}{1}, \frac{1}{8}, \frac{1}{27}, \frac{1}{64}, \frac{1}{125}, \dots \\ Wehavetofindthegeneraltermforthesequence. \end{array}$

Step 2. Observing the pattern

Now,

$\begin{matrix} n : & 1 & 2 & 3 & 4 & 5 & ... & n \\ Terms: & \frac{1}{1} & \frac{1}{8} & \frac{1}{27} & \frac{1}{64} & \frac{1}{125} & ... \\ Pattern & \frac{1}{1^{3}} & \frac{1}{2^{3}} & \frac{1}{3^{3}} & \frac{1}{4^{3}} & \frac{1}{5^{3}} & ... & \frac{1}{n^{3}} \end{matrix}$

Step 3. Conclusion

From the above table, we can observe that the general terms will be $a_{n} = \frac{1}{n^{3}} .$

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In the following exercise, find a general term for the sequence whose first five terms are shown.
$\frac{1}{1}, \frac{1}{8}, \frac{1}{27}, \frac{1}{64}, \frac{1}{125}, \dots$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Observing the pattern

Step 3. Conclusion

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In the following exercise, find a general term for the sequence whose first five terms are shown. 11,18,127,164,1125,…

Short Answer

Step by step solution

Step 1. Given information

Step 2. Observing the pattern

Step 3. Conclusion

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Applied Mathematics

Geometry

Statistics

Pure Maths

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

In the following exercise, find a general term for the sequence whose first five terms are shown.
$\frac{1}{1}, \frac{1}{8}, \frac{1}{27}, \frac{1}{64}, \frac{1}{125}, \dots$