Chapter 8.1, Problem 61E

To determine

To write: the first five terms of the sequence defined recursively and write the pattern for nth term.

Answer to Problem 61E

The first five terms are

  $81,27,9,3,1$ .

The pattern for nth term is $a_n=\frac{243}{3^n}$ .

Explanation of Solution

Given information:

Given sequence

  $a_1=81,a_{k+1}=\frac{1}{3}{a}_k$

Calculation:

Since the first term of the sequence is $a_1=81$ therefore, the subsequent terms can be computed from the recursive formula, $a_{k+1}=\frac{1}{3}{a}_k$ as follows,

  $\begin{array}{c}{a}_2=\frac{1}{3}{a}_1\\ =\frac{1}{3}\times 81\\ =27\\ a_3=\frac{1}{3}{a}_2\\ =\frac{1}{3}\times 27\\ =9\\ a_4=\frac{1}{3}{a}_3\\ =\frac{1}{3}\times 9\\ =3\\ a_5=\frac{1}{3}{a}_4\\ =\frac{1}{3}\times 3\\ =1\end{array}$

Hence the first five terms are

  $81,27,9,3,1$ .

Now, the pattern for nth term can be obtained as

Consider the sequence $81,27,9,3,1$

For $n=1,2,3,4,5,\mathrm{...}n$

The terms are $81,27,9,3,\mathrm{1...}{a}_n$

Rewrite the terms as shown below,

  $\begin{array}{c}81,27,9,3,\mathrm{1...}{a}_n=\frac{243}{3},\frac{243}{9},\frac{243}{27},\frac{243}{81},\frac{243}{243}\mathrm{...}{a}_n\\ =\frac{243}{3},\frac{243}{3^2},\frac{243}{3^3},\frac{243}{3^4},\frac{243}{3^5}\mathrm{...}{a}_n\end{array}$

. Therefore , the general term is $a_n=\frac{243}{3^n}$ .