Chapter 8.1, Problem 25E

(a)

To determine

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

Answer to Problem 25E

Five terms of this sequence are $3,15,35,63,99$ respectively.

Explanation of Solution

Given information:

Given sequence

  $a_n=\left(2n-1\right)\left(2n+1\right)$

Calculation:

Use the table feature of graphing utility to find the first 5 terms of sequence by using following steps:

Press Mode key, select seq in the fourth line and hit Enter

  

Press $Y=$ and Enter the sequence,

  

Ste the table by pressing TBLSET and the press TABLE,

  

(b)

To determine

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

Answer to Problem 25E

Five terms of this sequence are $3,15,35,63,99$ respectively.

Explanation of Solution

Calculation:

Given sequence

  $a_n=\left(2n-1\right)\left(2n+1\right)$

Value of $n$ begins with1, for five terms of this sequence put $n=1,2,3,4,5$ . So,

  $\begin{array}{c}{a}_1=\left(2\times 1-1\right)\left(2\times 1+1\right)\\ =1\times 3\\ =3\\ a_2=\left(2\times 2-1\right)\left(2\times 2+1\right)\\ =3\times 5\\ =15\\ a_3=\left(2\times 3-1\right)\left(2\times 3+1\right)\\ =5\times 7\\ =35\\ a_4=\left(2\times 4-1\right)\left(2\times 4+1\right)\\ =7\times 9\\ =63\\ a_5=\left(2\times 5-1\right)\left(2\times 5+1\right)\\ =9\times 11\\ =99\end{array}$

Hence, five terms of this sequence are $3,15,35,63,99$ respectively. So, it is clear that values obtained from both methods, are same.