Chapter 7.10, Problem 14PPS

To determine

To Find: The first five terms of each sequence $a_1=13,a_n=-2a_{n-1}-3,n\ge 2$

Answer to Problem 14PPS

First five terms of the sequence are $a_1=13,a_2=-29,a_3=55,a_4=-113\&a_5=223$

Explanation of Solution

Given information:

  $a_1=13,a_n=-2a_{n-1}-3,n\ge 2$

Concept and Formula Used:

To find the terms put the value of ‘n’.

Calculation:

It is given that

  $\begin{array}{l}{a}_1=13,a_n=-2a_{n-1}-3,n\ge 2\\ \text{Put }n=2\\ a_2=-2a_{2-1}-3\\ a_2=-2a_1-3\\ \text{As }{a}_1=13\\ a_2=-2\left(13\right)-3\\ a_2=-26-3=-29\\ \text{Now put }n=3\text{ in }{a}_n=-2a_{n-1}-3,n\ge 2\\ a_3=-2a_{3-1}-3\\ a_3=-2a_2-3\\ \text{As }{a}_2=-29\\ a_3=-2\left(-29\right)-3\\ a_3=58-3=55\\ \text{Now put }n=4\text{ in }{a}_n=-2a_{n-1}-3,n\ge 2\\ a_4=-2a_{4-1}-3\\ a_4=-2a_3-3\\ \text{As }{a}_3=55\\ a_4=-2\left(55\right)-3\\ a_4=-110-3=-113\\ \text{Now put }n=5\text{ in }{a}_n=-2a_{n-1}-3,n\ge 2\\ a_5=-2a_{5-1}-3\\ a_5=-2a_4-3\\ \text{As }{a}_4=-113\\ a_5=-2\left(-113\right)-3\\ a_5=226-3=223\end{array}$

Conclusion:

First five terms of the sequence are $a_1=13,a_2=-29,a_3=55,a_4=-113\&a_5=223$