Chapter 8.1, Problem 68E

(a)

To determine

To write the first five terms of the given sequence using table feature of a graphing utility.

Answer to Problem 68E

First five terms of this sequence are $-1,0.16667,-\frac{1}{120},-\frac{1}{5040},-\frac{1}{362880}$ respectively.

Explanation of Solution

Given information:

Given sequence

  $a_n=\frac{{\left(-1\right)}^{2n+1}}{\left(2n+1\right)!}$

Calculation:

To have the first five terms by graphical utility, firstly click on $Y=$ button of Ti-83 calculator and then write down the provided function of the problem, as shown below:

  

Now, Set the table setup. For this, click on $2^{nd}$ and then click on WINDOW button. After this a new screen appears, In this, set the indpnt as Ask. The screenshot is given below:

  

After this, click on $2^{nd}$ and then click on GRAPH button. After this, a new screen appears. In this, input the degrees to have the required table. The screenshot is given as below:

  

(b)

To determine

To write the first five terms of the given sequence algebraically.

Answer to Problem 68E

First five terms of this sequence are $1,-0.5,0.04167,-\frac{1}{720},\frac{1}{40302}$ respectively.

Explanation of Solution

Calculation:

Given sequence

  $a_n=\frac{{\left(-1\right)}^{2n+1}}{\left(2n+1\right)!}$

Value of $n$ begins with1, for five terms of this sequence put $n=0,1,2,3,4$ . So,

  $\begin{array}{c}{a}_0=\frac{{\left(-1\right)}^{0+1}}{\left(0+1\right)!}\\ =-1\\ a_1=\frac{{\left(-1\right)}^{2+1}}{\left(2+1\right)!}\\ =-\frac{1}{3!}\\ =-\frac{1}{3\times 2\times 1}\\ =-\frac{1}{6}\\ =-0.16667\\ a_2=\frac{{\left(-1\right)}^{4+1}}{\left(2\times 2+1\right)!}\\ =-\frac{1}{5!}\\ =\frac{1}{5\times 4\times 3\times 2\times 1}\\ =\frac{1}{120}\\ =0.04167\\ a_3=\frac{{\left(-1\right)}^{6+1}}{\left(2\times 3+1\right)!}\\ =-\frac{1}{7!}\\ =-\frac{1}{7\times 6\times 5\times 4\times 3\times 2\times 1}\\ =-\frac{1}{5040}\\ a_4=\frac{{\left(-1\right)}^{8+1}}{\left(2\times 4+1\right)!}\\ =-\frac{1}{9!}\\ =-\frac{1}{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\\ =-\frac{1}{362880}\end{array}$

Hence, five terms of this sequence are $-1,0.16667,-\frac{1}{120},-\frac{1}{5040},-\frac{1}{362880}$ respectively. So, it is clear that values obtained from both methods, are same.