Chapter 9.1, Problem 54E

To determine

To find the first five terms of the sequence defined recursively as $a_{k+1}=\left(-2\right)a_k$ where $a_1=14$ .

Answer to Problem 54E

  $a_1=14$ , $a_2=-28$ , $a_3=56$ , $a_4=-112$ , $a_5=224$

Explanation of Solution

Given information :

Let recursive formula be defined as: $a_{k+1}=\left(-2\right)a_k$ where $a_1=14$ .

Concept used:

A sequence is a function whose domain is the set of positive integers. The function values $a_1,a_2,a_3,a_4,...,a_n,...$ are the terms of the sequence.

Calculations:

For $k=1$ , the recursive formula will be $a_{1+1}=a_2=\left(-2\right)a_1$

Here, substitute $a_1=14$ and get,

  $a_2=a_{1+1}=\left(-2\right)a_1=-2\left(14\right)=-28$

Similarly,

For $k=2$ and $a_2=-28$ :

  $a_3=a_{2+1}=\left(-2\right)a_2=-2\left(-28\right)=56$

For $k=3$ and $a_3=56$ :

  $a_4=a_{3+1}=\left(-2\right)a_3=-2\left(56\right)=-112$

For $k=4$ and $a_4=-112$ :

  $a_5=a_{4+1}=\left(-2\right)a_4=-2\left(-112\right)=224$