Chapter 9, Problem 28RE

Solution

To calculate: The first 6 terms and the 40^th term the sequence $b_k=\frac{{\left(-2\right)}^k}{k+1}$ .

The obtained terms is $-1,\frac{4}{3},-2,\frac{16}{5},-\frac{16}{3},$ and $2.6\cdot {10}^{10}$ .

Given Information:

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

Calculation:

Consider the given information,

  $b_k=\frac{{\left(-2\right)}^k}{k+1}$

Substitute 1 for $k$ in the given sequence.

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

Substitute 2 for $k$ in the given sequence.

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

Substitute 3 for $k$ in the given sequence.

  $\begin{array}{c}{b}_3=\frac{{\left(-2\right)}^3}{3+1}\\ =-\frac{8}{4}\\ =-2\end{array}$

Substitute 4 for $k$ in the given sequence.

  $\begin{array}{c}{b}_4=\frac{{\left(-2\right)}^4}{4+1}\\ =\frac{16}{5}\end{array}$

Substitute 5 for $k$ in the given sequence.

  $\begin{array}{c}{b}_5=\frac{{\left(-2\right)}^5}{5+1}\\ =-\frac{32}{6}\\ =-\frac{16}{3}\end{array}$

Substitute 6 for $k$ in the given sequence.

  $\begin{array}{c}{b}_{40}=\frac{{\left(-2\right)}^{40}}{40+1}\\ =\frac{2^{40}}{41}\\ =2.6\cdot {10}^{10}\end{array}$

Substitute 40 for $k$ in the given sequence.

  $\begin{array}{c}{a}_{40}=\frac{{40}^2-1}{40+1}\\ =\frac{1599}{41}\\ =39\end{array}$

Hence, the obtained terms is $-1,\frac{4}{3},-2,\frac{16}{5},-\frac{16}{3},$ and $2.6\cdot {10}^{10}$ .