Chapter 11, 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

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

Answer to Problem 1MCCP

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

Explanation of Solution

Given:

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

Formula used:

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

Calculation:

Consider the provided general term of sequence,

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

In order to find the first four terms of the sequence, substitute $n=1,2,3,4$ in the general term.

For first term, put $n=1$,

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

For second term, put $n=2$,

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

For third term, put $n=3$,

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

For fourth term, put $n=4$,

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

For fifth term, put $n=5$,

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

Thus, first five terms $a_1,a_2,a_3,a_4,a_5$ are $1,-2,\frac{3}{2},-\frac{2}{3},\frac{5}{24}$ respectively.