Chapter 8.5, Problem 56E

To determine

To give:an example of the given sequence. And write a recursive rule for the sequence and find its first eight terms.

Answer to Problem 56E

  $1,2,1,-1,-3,1,6,1$ .

Explanation of Solution

Given:

A sequence in which each term after the third term is a function of the three terms preceding it.

Calculation:

An example of a sequence in which each term after the third term is a function of the three terms preceding it.

Let us consider the sequence:

  $\begin{array}{l}{a}_1=1\\ a_2=2\\ a_3=2\\ a_n=a_{n-3}-2a_{n-2}+a_{n-1}\end{array}$

Now write the first eight terms

  $\begin{array}{l}{a}_4=a_{4-3}-2a_{4-2}+a_{4-1}\\ =a_1-2a_2+a_3=1-2\times 2+2=-1\\ a_5=a_{5-3}-2a_{5-2}+a_{5-1}\\ =a_2-2a_3+a_4=2-2\times 2+2=-3\\ a_6=a_{6-3}-2a_{6-2}+a_{6-1}\\ =a_3-2a_4+a_5=2-2\times \left(-1\right)-3=1\\ a_7=a_{7-3}-2a_{7-2}+a_{7-1}\\ =a_4-2a_5+a_6=-1-2\times \left(-3\right)+1=6\\ a_8=a_{8-3}-2a_{8-2}+a_{8-1}\\ =a_5-2a_6+a_7=-3-2\times 1+6=1\end{array}$

Thus, the first eight terms of the given sequence are $1,2,1,-1,-3,1,6,1$ .