Chapter 9, Problem 2RE

To determine

To find the first four terms and the 40^th term of the given sequence.

Answer to Problem 2RE

First four terms and 40^th team is,

  $-3,-6,-12,-24\text{and}-{3.2}^{39}.$

Explanation of Solution

Given information:

Given sequence is,

  $a_1=-3,an=2a_{n-1}$ For all $n\ge 2$

CALCULATION:

Given sequence is,

  $a_1=-3,a_n=2a_{n-1}$

To find the first term of the above given sequence. Find $a_1$ where $n=1$ .

  $a_1=-3.$

To find the 2^nd term of the above given sequence. Find $a_2$ where $n=2$ .

  $a_2=2a_{2-1}=2a_1=2\times \left(-3\right)=-6\text{}\left[\text{since, }{a}_1=-3,n=2\right]$

To find the third term of the above given sequence. Find $a_3$ where $n=3$ .

  $a_3=2a_{3-1}=2a_2=2\times \left(-6\right)=-12\text{}\left[\text{since, }{a}_2=-6,n=3\right]$

To find the fourth term of the above given sequence. Find $a_4$ where $n=4$

  $a_4=2a_{4-1}=2a_3=2\times \left(-12\right)=-24\text{}\left[\text{since, }{a}_3=-12,n=4\right]$

To find the 40^th term of the above given sequence. Find $a_{40}$ where $n=40$

  $a_{40}=2a_{40-1}=2a_{39}=3\cdot 2^{39}\left[\text{since, }n=40\right]$

Hence,

The first four term and 40^th term is $-3,-6,-12,-24\text{and}-3\cdot 2^{39}$