Chapter 8.5, Problem 67E

a.

To determine

To write: the first five terms of the sequence.

Answer to Problem 67E

  $3.10,21,36,55$ .

Explanation of Solution

Given:

The rule for a recursive sequence is as follows:

  $\begin{array}{l}f\left(1\right)=3,f\left(2\right)=10\\ f\left(n\right)=4+2f\left(n-1\right)-f\left(n-2\right)\end{array}$ .

Calculation:

According to the question, the rule for a recursive sequence is as follows:

  $\begin{array}{l}f\left(1\right)=3,f\left(2\right)=10\\ f\left(n\right)=4+2f\left(n-1\right)-f\left(n-2\right)\end{array}$ .

Now, to find the value of the sequence $a_3,a_4,a_5$ . Then

  $\begin{array}{l}f\left(3\right)=4+2f\left(3-1\right)-f\left(3-2\right)\\ f\left(3\right)=4+2f\left(2\right)-f\left(1\right)\\ f\left(3\right)=4+2\left(10\right)-\left(3\right)=21\\ f\left(4\right)=4+2f\left(4-1\right)-f\left(4-2\right)\\ f\left(4\right)=4+2f\left(3\right)-f\left(2\right)\\ f\left(4\right)=4+2\left(21\right)-\left(10\right)=36\\ f\left(5\right)=4+2f\left(5-1\right)-f\left(5-2\right)\\ f\left(5\right)=4+2f\left(4\right)-f\left(3\right)\\ f\left(5\right)=4+2\left(36\right)-\left(21\right)=55\end{array}$

Hence, the first five terms of the sequence are $3,10,21,36,55$ .

b.

To determine

To find: the pattern by using the finite difference.

Answer to Problem 67E

Quadratic $f\left(3\right)+f\left(1\right)=2f\left(2\right)+4$ .

Explanation of Solution

Given:

The rule for a recursive sequence is as follows:

  $\begin{array}{l}f\left(1\right)=3,f\left(2\right)=10\\ f\left(n\right)=4+2f\left(n-1\right)-f\left(n-2\right)\end{array}$ .

Calculation:

From part (a) the sequences are $3.10,21,36,55$ .

  $\begin{array}{l}f\left(1\right)=3\\ f\left(2\right)=10\\ f\left(3\right)=21\\ f\left(4\right)=36\\ \because f\left(2\right)-f\left(1\right)=7\\ f\left(3\right)-f\left(2\right)=11\\ f\left(4\right)-f\left(3\right)=15\\ \therefore \left[\left(f\left(3\right)-f\left(2\right)\right)-\left(f\left(2\right)-f\left(1\right)\right)\right]=4\\ \left[\left(f\left(4\right)-f\left(3\right)\right)-\left(f\left(3\right)-f\left(2\right)\right)\right]=4\\ \therefore f\left(3\right)+f\left(1\right)=2f\left(2\right)+4\end{array}$

Thus, the pattern is $f\left(3\right)+f\left(1\right)=2f\left(2\right)+4$ and the nature of relation is quadratic.

c.

To determine

To write: an explicit rule for the given sequence.

Answer to Problem 67E

  $f\left(n\right)=2{\left(n\right)}^2+n$ .

Explanation of Solution

Given:

The rule for a recursive sequence is as follows:

  $\begin{array}{l}f\left(1\right)=3,f\left(2\right)=10\\ f\left(n\right)=4+2f\left(n-1\right)-f\left(n-2\right)\end{array}$ .

Calculation:

From part (a) the sequences are $3.10,21,36,55$ .

  $\begin{array}{l}f\left(1\right)=3=2{\left(1\right)}^2+1\\ f\left(2\right)=10=2{\left(2\right)}^2+2\\ f\left(3\right)=21=2{\left(3\right)}^2+3\\ \mathrm{.................................}\\ \mathrm{.................................}\\ \mathrm{.................................}\\ f\left(n\right)=2{\left(n\right)}^2+n\end{array}$

Thus, the explicit rule for the sequence is $f\left(n\right)=2{\left(n\right)}^2+n$ .