Question

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

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

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

Step 2. Observing the pattern

Now,

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

Step 3. Conclusion

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