Chapter 8.5, Problem 61E

a.

To determine

To write: a recursive rule for the number of trees on the tree farm at the beginning of the nth year.

Answer to Problem 61E

  $a_{n-1}={\left(0.9\right)}^{n-2}{a}_1+800\left[1-{\left(0.9\right)}^{n-2}\right]$ .

Explanation of Solution

Given:

A tree family has 9000 trees increases $10\%$ every year and 800 seedlings are planted.

Calculation:

Let $a_n$ is the number of trees at the end of nth year. Then

  $\begin{array}{l}{a}_1=9000\\ a_2=\left(1-0.10\right)a_1+800=0.9a_1+800\\ a_3=\left(1-0.10\right)a_2+800\\ =0.9\left(0.9a_1+800\right)+800\\ ={\left(0.9\right)}^2{a}_1\left(0.9\right)800+800\end{array}$

Therefore, the number of trees at the beginning of the nth year is the number of trees at the end of $\left(n-1\right)th$ year. Therefore,

  $\begin{array}{l}{a}_{n-1}={\left(0.9\right)}^{n-2}{a}_1+800\left[{\left(0.9\right)}^{n-3}+\mathrm{.....}+0.9+1\right]\left[\left(800\right)\right]\\ a_{n-1}={\left(0.9\right)}^{n-2}{a}_1+800\times \frac{1-{\left(0.9\right)}^{n-2}}{1-0.9}\\ a_{n-1}={\left(0.9\right)}^{n-2}{a}_1+800\left[1-{\left(0.9\right)}^{n-2}\right]\end{array}$

Hence, the amount after the fourth payment is $a_{n-1}={\left(0.9\right)}^{n-2}{a}_1+800\left[1-{\left(0.9\right)}^{n-2}\right]$ .

b.

To determine

To check: that happens in number of trees after an extended period of time.

Answer to Problem 61E

  $a_{n-1}=8000$ .

Explanation of Solution

Given:

The borrow amount of $\$10,000$ at $11.5\%$ annual interest compounded monthly for 7 years.

Calculation:

From part (a)

  $a_{n-1}={\left(0.9\right)}^{n-2}{a}_1+800\left[1-{\left(0.9\right)}^{n-2}\right]$

When n increases lot, then, ${\left(0.9\right)}^{n-2}\to 0$

Therefore, the number of trees is $a_{n-1}=8000$