Chapter 9.1, Problem 53E

To determine

To find the first five terms of the sequence defined recursively as $a_{k+1}=\frac{1}{3}{a}_k$ where $a_1=81$ .

Answer to Problem 53E

  $a_1=81$ , $a_2=27$ , $a_3=9$ , $a_4=3$ , $a_5=1$

Explanation of Solution

Given information :

Let recursive formula be defined as: $a_{k+1}=\frac{1}{3}{a}_k$ where $a_1=81$ .

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 $a_{k+1}=\frac{1}{3}{a}_k$

Here, substitute $a_1=81$ and get,

  $a_2=a_{1+1}=\frac{1}{3}{a}_1=\frac{1}{3}\times 81=27$

Similarly,

For $k=2$ and $a_2=27$ :

  $a_3=a_{2+1}=\frac{1}{3}{a}_2=\frac{1}{3}\times 27=9$

For $k=3$ and $a_3=9$ :

  $a_4=a_{3+1}=\frac{1}{3}{a}_3=\frac{1}{3}\times 9=3$

For $k=4$ and $a_4=3$ :

  $a_5=a_{4+1}=\frac{1}{3}{a}_4=\frac{1}{3}\times 3=1$