Chapter 9.3, Problem 10E

Solution

To Calculate: The first 4 terms and the 8th term of the recursively defined sequence $c_1=-2,c_2\text{=}3\text{ and }{c}_k=c_{k-2}+c_{k-1}\text{ for }k\ge 3$ .

The first 4 terms and the 8th terms are $-2,3,1,4,\text{ and 2}3$ respectively.

Given information:

The givens sequence is $c_1=-2,c_2\text{=}3\text{ and }{c}_k=c_{k-2}+c_{k-1}\text{ for }k\ge 3$

Formula used:

Substitute the given values into the given sequence.

Calculation:

Consider the given sequence.

The first term is $-2$ and second term is 3.

Put $k=3$ into the given sequence.

  $\begin{array}{c}{c}_3=c_{3-2}+c_{3-1}\\ c_3=c_1+c_2\\ =-2+3\\ =1\end{array}$

For $k=4$

  $\begin{array}{c}{c}_4=c_{4-2}+c_{4-1}\\ c_4=c_2+c_3\\ =3+1\\ =4\end{array}$

For $k=5$

  $\begin{array}{c}{c}_5=c_{5-2}+c_{5-1}\\ c_5=c_3+c_4\\ =1+4\\ =5\end{array}$

For $k=6$

  $\begin{array}{c}{c}_6=c_{6-2}+c_{6-1}\\ c_6=c_4+c_5\\ =4+5\\ =9\end{array}$

For $k=7$

  $\begin{array}{c}{c}_7=c_{7-2}+c_{7-1}\\ c_7=c_5+c_6\\ =5+9\\ =14\end{array}$

For $k=8$

  $\begin{array}{c}{c}_8=c_{8-2}+c_{8-1}\\ c_8=c_6+c_7\\ =9+14\\ =23\end{array}$

The first 4 terms and the 8th terms are $-2,3,1,4,\text{ and 2}3$ respectively.