Chapter 7.5, Problem 48PS

Solution

i.

To state: A recursive equation that shows $a_n$ is related to $a_{n-1}$ , $a_n$ is the balance after n years after retiring if you plan to withdraw $30,000 at the beginning of each year for 20 years after you retire. Based on the type of investment you are making, you can expect to earn an annual return of 8% on your savings after you retire.

The resultant recursive rule is $a_n=1.08a_{n-1}-30000$ .

Given information:

You plan to withdraw $30,000 at the beginning of each year for 20 years after you retire. Based on the type of investment you are making, you can expect to earn an annual return of 8% on your savings after you retire.

Explanation:

  $a_n$ is the balance after n years after retiring.

$30,000 are to be withdrawn at the beginning of each year for 20 years and 8% annual return is expected on your savings after you retire.

Now form the recursive rule that shows the relation of $a_n$ to $a_{n-1}$ is,

  $\begin{array}{c}{a}_n=0.08a_{n-1}-30000+a_{n-1}\\ =1.08a_{n-1}-30000\end{array}$

Therefore, the recursive rule is $a_n=1.08a_{n-1}-30000$ .

ii.

To state: The minimum amount of money you should have in your account when you retire if $a_{20}=0$ .

The resultant amount is $294544.42.

Given information:

It is given that $a_{20}=0$ .

Explanation:

The recursive rule from part (a) is: $a_n=1.08a_{n-1}-30000$

Solve it for $a_{n-1}$ ,

  $\begin{array}{c}{a}_n=1.08a_{n-1}-30000\\ a_n+30000=1.08a_{n-1}\\ a_{n-1}=\frac{a_n+30000}{1.08}\end{array}$

It is given that $a_{20}=0$ , the value $a_0$ is needed to tell the minimum amount of money needed to be in account.

  $\begin{array}{l}{a}_{19}=\frac{a_{20}+30000}{1.08},a_{19}=27777.78\\ a_{18}=\frac{a_{19}+30000}{1.08},a_{18}=53497.9\\ a_{17}=\frac{a_{18}+30000}{1.08},a_{17}=77312.9\\ a_{16}=\frac{a_{17}+30000}{1.08},a_{16}=99363.8\\ a_{15}=\frac{a_{16}+30000}{1.08},a_{15}=119781.3\\ a_{14}=\frac{a_{15}+30000}{1.08},a_{14}=138686.4\\ a_{13}=\frac{a_{12}+30000}{1.08},a_{13}=156191.1\\ a_{12}=\frac{a_{13}+30000}{1.08},a_{12}=172399.16\\ a_{11}=\frac{a_{12}+30000}{1.08},a_{11}=187406.6\end{array}$

Similarly, find other terms,

  $\begin{array}{c}{a}_{10}=201302.4,a_9=214168.9\\ a_8=226082.3,a_7=217113.3\\ a_6=247327.1,a_5=256784.4\\ a_4=265541.07,a_3=273649.14\\ a_2=281156.61,a_1=288107.976\\ a_0=294544.42\end{array}$

Therefore, the minimum amount of money should have been in the account is $294544.42.