Chapter 8.5, Problem 33E

To determine

To write: a recursive rule for the given sequence.

Answer to Problem 33E

  $a_n=11a_{n-1}$ .

Explanation of Solution

Given:

The given sequence is $a_n=12{\left(11\right)}^{n-1}$ .

Calculation:

The given sequence is $a_n=12{\left(11\right)}^{n-1}$ .

Now, substitutes the value of $n=1,2,3,\mathrm{.....}\text{ and so on}$ . Therefore,

  $\begin{array}{l}{a}_n=12{\left(11\right)}^{n-1}\{\text{given sequences}}\\ a_1=12{\left(11\right)}^{1-1}\{\text{substitues }n=1\text{}}\\ a_1=12\{\because {11}^0=1\text{}}\\ a_2=12{\left(11\right)}^{2-1}\{\text{substitues }n=2\text{}}\\ a_2=12\times 11=132\end{array}$

  $\begin{array}{l}\because a_n=a_{n-1}+d\\ \therefore a_2=11a_{2-1}\\ \Rightarrow 132=11a_{2-1}=132\left\{true\right\}\\ \therefore a_n=11a_{n-1}\end{array}$

Hence, the recursive rule for the given sequence is $a_n=11a_{n-1}$ .