Chapter 13, Problem 1RE

To determine

The first five terms of the sequence $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$.

Answer to Problem 1RE

The first five terms of the sequence $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$ are $-\frac{4}{3},\frac{5}{4},-\frac{6}{5},\frac{7}{6},-\frac{8}{7}$.

Explanation of Solution

The given sequence is $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$.

The first five terms of the sequence are calculated as follows.

Substitute $n=1$ in the sequence $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$ and obtain the first term as,

$\begin{array}{c}{a}_1={\left(-1\right)}^1\left(\frac{1+3}{1+2}\right)\\ =-\frac{4}{3}\end{array}$

Substitute $n=2$ in the sequence $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$ and obtain the second term as,

$\begin{array}{c}{a}_2={\left(-1\right)}^2\left(\frac{2+3}{2+2}\right)\\ =\frac{5}{4}\end{array}$

Substitute $n=3$ in the sequence $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$ and obtain the third term as,

$\begin{array}{c}{a}_3={\left(-1\right)}^3\left(\frac{3+3}{3+2}\right)\\ =-\frac{6}{5}\end{array}$

Substitute $n=4$ in the sequence $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$ and obtain the fourth term as,

$\begin{array}{c}{a}_4={\left(-1\right)}^4\left(\frac{4+3}{4+2}\right)\\ =\frac{7}{6}\end{array}$

Substitute $n=5$ in the sequence $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$ and obtain the fifth term as,

$\begin{array}{c}{a}_5={\left(-1\right)}^5\left(\frac{5+3}{5+2}\right)\\ =-\frac{8}{7}\end{array}$

Therefore, the first five terms of the sequence $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$ are $-\frac{4}{3},\frac{5}{4},-\frac{6}{5},\frac{7}{6},-\frac{8}{7}$.