Chapter 7.10, Problem 4AGP

To determine

The recursive formula from the given explicit formula.

Answer to Problem 4AGP

The recursive formula is $a_1=4$ , $a_n=3a_{n-1}$ , $n\ge 2$

Explanation of Solution

Given information:The explicit formula given $a_n=4{(3)}^{n-1}$ .

Calculation:

The formula given $a_n=4{(3)}^{n-1}$ .

By plugging in n=1, first term $a_1=4{(3)}^{1-1}=4{(3)}^0=4(1)=4$ .

When n=2, second term $a_2=4{(3)}^{2-1}=4{(3)}^1=4(3)=12$ .

When n=3, third term $a_3=4{(3)}^{3-1}=4{(3)}^2=4(9)=36$ .

When n=4, fourth term $a_4=4{(3)}^{4-1}=4{(3)}^3=4(27)=108$ .

So the sequence becomes 4, 12, 36, 108,….

We can write it as,

  $\begin{array}{l}{a}_2=12=4\times 3=a_1\times 3=3a_1\\ a_3=36=12\times 3=a_2\times 3=3a_2\\ a_4=108=36\times 3=a_3\times 3=3a_3\end{array}$

  $\therefore$ we can write, $a_n=3a_{n-1}$ .

So the recursive formula we can write, $a_1=4$ , $a_n=3a_{n-1}$ , $n\ge 2$ .