Chapter 12.1, Problem 32AYU

In Problems 29-36, the given pattern continues. Write down the nth term of a sequence $\left\{a_n\right\}$ suggested by the pattern.

$1,\frac{1}{2},\text{ 3, }\frac{1}{4},\text{ 5, }\frac{1}{6},\text{ 7, }\frac{1}{8},\mathrm{...}$

To determine

To find: The nth term of a sequence $\left\{a_n\right\}$ suggested by the pattern.

Answer to Problem 32AYU

$\left\{a_n\right\}{\text{ = n}}^{{\left(-1\right)}^{\text{n }-1}}$

Explanation of Solution

Given:

$1,\frac{1}{2},3,\frac{1}{4},5,\frac{1}{6},7,\frac{1}{8},\dots .$

Calculation:

The given series is $1,\frac{1}{2},3,\frac{1}{4},5,\frac{1}{6},7,\frac{1}{8},\dots .$

The first term is 1.

The second term is $\frac{1}{2}$.

The nth term will be ${\text{n}}^{{\left(-1\right)}^{\text{n }-1}}$.

$\left\{a_n\right\}{\text{ = n}}^{{\left(-1\right)}^{\text{n }-1}}$