Chapter 13.7, Problem 58STP

a.

To determine

To find: Recursive equation for the balance $b_n$ at the end of $n$ years.

Answer to Problem 58STP

Recursive equation for the balance amount after $n$ years is

  $b_n=b_{n-1}\left(1.03\right)-10$

Explanation of Solution

Given information:

Rita has deposited $\$1000$ in her bank account

  $b_0=\$1000$

Annual fee $=\$10$

The bank posts interest to her account in $3\%$ amount of the balance

Given

  $b_0=\$1000$

Every year

The bank posts interest to her account in $3\%$ amount of the balance

It means

Total amount after one year

Will be

Amount $=1000+\frac{3}{100}\times 1000$

  $\Rightarrow$ Amount $=1000\left(1+\frac{3}{100}\right)$

  $\Rightarrow$ Amount $=1000\left(1+0.03\right)$

  $\Rightarrow$ Amount $=1000\left(1.03\right)$

There will annual fees of $\$10$ every year

Therefore,

Amount after one year wil be

  $\Rightarrow b_1=1000\left(1.03\right)-10$

  $\Rightarrow b_1=b_0\left(1.03\right)-10$

Amount after two years will be

Amount $=\left(1000\left(1.03\right)-10\right)+\frac{3}{100}\times \left(1000\left(1.03\right)-10\right)$

  $\Rightarrow$ Amount $=\left(1000\left(1.03\right)-10\right)\left(1+\frac{3}{100}\right)$

  $\Rightarrow$ Amount $=\left(1000\left(1.03\right)-10\right)\left(1+0.03\right)$

  $\Rightarrow$ Amount $=\left(1000\left(1.03\right)-10\right)\left(1.03\right)$

Therefore,

Total amount after two years incuding annual fees is

  $\Rightarrow b_2=\left(1000\left(1.03\right)-10\right)\left(1.03\right)-10$

  $\Rightarrow b_2=b_1\left(1.03\right)-10$

Therefore,

On comparing $b_1,b_2$

We can write a general equation that is

  $\Rightarrow b_n=b_{n-1}\left(1.03\right)-10$

Therefore,

Recursive equation for the balance amount after $n$ years is

  $b_n=b_{n-1}\left(1.03\right)-10$

b.

To determine

To find: Balance in the account after four years.

Answer to Problem 58STP

Balance in account after four years $=b_4=\$1083.67$

Explanation of Solution

Given information:

Rita has deposited $\$1000$ in her bank account

  $b_0=\$1000$

Annual fee $=\$10$

The bank posts interest to her account in $3\%$ amount of the balance

Recursive equation for the balance amount after $n$ years is

  $b_n=b_{n-1}\left(1.03\right)-10$

Substitute

  $n=1$

We get

  $\begin{array}{l}{b}_n=b_{n-1}\left(1.03\right)-10\\ \Rightarrow b_1=b_{1-1}\left(1.03\right)-10\\ \Rightarrow b_1=b_0\left(1.03\right)-10\\ \Rightarrow b_1=1000\left(1.03\right)-10\\ \Rightarrow b_1=1030-10\\ \Rightarrow b_1=1020\end{array}$

Substitute

  $n=2$

We get

  $\begin{array}{l}{b}_n=b_{n-1}\left(1.03\right)-10\\ \Rightarrow b_2=b_{2-1}\left(1.03\right)-10\\ \Rightarrow b_2=b_1\left(1.03\right)-10\\ \Rightarrow b_2=1020\left(1.03\right)-10\\ \Rightarrow b_2=1050.6-10\\ \Rightarrow b_2=1040.6\end{array}$

Substitute

  $n=3$

We get

  $\begin{array}{l}{b}_n=b_{n-1}\left(1.03\right)-10\\ \Rightarrow b_3=b_{3-1}\left(1.03\right)-10\\ \Rightarrow b_3=b_2\left(1.03\right)-10\\ \Rightarrow b_3=1040.6\left(1.03\right)-10\\ \Rightarrow b_3=1071.82-10\\ \Rightarrow b_3=1061.82\end{array}$

Substitute

  $n=4$

We get

  $\begin{array}{l}{b}_n=b_{n-1}\left(1.03\right)-10\\ \Rightarrow b_4=b_{4-1}\left(1.03\right)-10\\ \Rightarrow b_4=b_3\left(1.03\right)-10\\ \Rightarrow b_4=1061.82\left(1.03\right)-10\\ \Rightarrow b_4=1093.67-10\\ \Rightarrow b_4=1083.67\end{array}$

Therefore,

Balance in account after four years $=b_4=\$1083.67$