Chapter 8.5, Problem 8E

To determine

To write:the first six terms of the given sequence.

Answer to Problem 8E

  $1,-9,71,5031,25310951,6.4\times {10}^{14}$ .

Explanation of Solution

Given:

The givensequence is $a_n={\left(a_{n-1}\right)}^2-10$ .

  $a_1=1$

Calculation:

According to the given sequence, $a_n={\left(a_{n-1}\right)}^2-10$ .

Therefore, the first six terms of the given sequenceare $a_2,a_3,a_4,a_5,a_6$ .

Now, to find the first six terms of the given sequence is given by

  $\begin{array}{l}{a}_n={\left(a_{n-1}\right)}^2-10\{\text{given sequence}}\\ a_n={\left(a_{2-1}\right)}^2-10\{\text{put }n=2\text{}}\\ a_2={\left(a_1\right)}^2-10\\ a_2={\left(1\right)}^2-10=-9\{\text{substitutes }{a}_1=1\text{}}\\ a_3={\left(a_{3-1}\right)}^2-10\{\text{put }n=3\text{}}\\ a_3={\left(a_2\right)}^2-10\\ a_3={\left(-9\right)}^2-10=71\{\text{substitutes }{a}_2=-9\text{}}\end{array}$

  $\begin{array}{l}{a}_4={\left(a_{4-1}\right)}^2-10\{\text{put }n=4\text{}}\\ a_4={\left(a_3\right)}^2-10\\ a_4={\left(71\right)}^2-10=5031\{\text{substitutes }{a}_3=71\text{}}\end{array}$

  $\begin{array}{l}{a}_5={\left(a_{5-1}\right)}^2-10\{\text{put }n=5\text{}}\\ a_5={\left(a_4\right)}^2-10\\ a_5={\left(5031\right)}^2-10=25310951\{\text{substitutes }{a}_4\text{=5031}}\\ a_6={\left(a_{6-1}\right)}^2-10\{\text{put }n=5\text{}}\\ a_6={\left(a_5\right)}^2-10\\ a_6={\left(25310951\right)}^2-10\approx 6.4\times {10}^{14}\{\text{substitutes }{a}_5=25310951\text{}}\end{array}$

Hence, the first six terms of the given sequence are $1,-9,71,5031,25310951,6.4\times {10}^{14}$ .