Chapter 8, Problem 1MCCP

In Exercises 1-4, write the first five terms of each sequence. Assume that d represents the common difference of an arithmetic sequence and r represents the common ratio of a geometric sequence.

$a_n={(-1)}^{n+1}\frac{n}{(n-1)!}$

To determine

The firstfive terms of the geometric sequence.

Answer to Problem 1MCCP

Solution: $1,-2,\frac{3}{2},-\frac{2}{3},\frac{5}{24}$

Explanation of Solution

Given: $a_n={\left(-1\right)}^{n+1}\frac{n}{\left(n-1\right)!}$

Consider the provided expression;

$a_n={\left(-1\right)}^{n+1}\frac{n}{\left(n-1\right)!}$ (I)

Substitute $n=1$ in (I) to get the first term;

$\begin{array}{c}{a}_1={\left(-1\right)}^{1+1}\frac{1}{\left(1-1\right)!}\\ ={\left(-1\right)}^2\frac{1}{\left(0\right)!}\\ =\frac{1}{1}\\ =1\end{array}$

Substitute $n=2$ in (I) to get the second term;

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

Substitute $n=3$ in (I) to get the third term;

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

Substitute $n=4$ in (I) to get the fourth term;

$\begin{array}{c}{a}_4={\left(-1\right)}^{4+1}\frac{4}{\left(4-1\right)!}\\ ={\left(-1\right)}^5\frac{4}{\left(3\right)!}\\ =-\frac{2}{3}\end{array}$

Substitute $n=5$ in (I) to get the fifth term;

$\begin{array}{c}{a}_5={\left(-1\right)}^{5+1}\frac{5}{\left(5-1\right)!}\\ ={\left(-1\right)}^6\frac{5}{\left(4\right)!}\\ =\frac{5}{24}\end{array}$

Conclusion: The first five terms of the geometric sequence are $1,-2,\frac{3}{2},-\frac{2}{3},\frac{5}{24}$.