Chapter 6.6, Problem 39E

To determine

To write a recursive rule for the sequence $1,3,4,7,11,\mathrm{......}$ . And write next two terms of the sequence .

Answer to Problem 39E

The recursive rule is $a_1=1,a_2=3,a_n=a_{n-1}+a_{n-2}$ and next two terms of the given sequence are $18,29$

Explanation of Solution

Given information:

Given sequence $1,3,4,7,11,\mathrm{......}$ .

The given $1,3,4,7,11,\mathrm{......}$ is not arithmetic sequence because there does not exists common difference $\begin{array}{l}d=a_2-a_1=3-1=2andalso{a}_3-a_2=4-3=1\\ \Rightarrow 2\ne 1\end{array}$ .

Similar way the given sequence is not a geometric sequence as well because there does not common ratio

  $\begin{array}{l}r=\frac{a_2}{a_1}=\frac{3}{1}=3andalso\frac{a_3}{a_2}=\frac{4}{3}\\ \Rightarrow 3\ne \frac{4}{3}\end{array}$

Thus, now find $a_1+a_2=1+3=4$ is the third term ( $a_3$ )

Similar way, $a_2+a_3=3+4=7$ is the fourth term ( $a_4$ )

Thus, each term is the sum of the two previous terms,

Therefore, a recursive equation for the given sequence is $a_1=1,a_2=3,a_n=a_{n-1}+a_{n-2}$

Next two term of the given sequence are $18,29.$

Hence, recursive rule is $a_1=1,a_2=3,a_n=a_{n-1}+a_{n-2}$ and next two terms of the given sequence are $18,29$