Chapter 5.5, Problem 43PS

Solution

a.

To prove: The formulas are equivalent by simplifying the first formula.

Given information:

Formula- $1$ :

  $M=\frac{Pi}{1-{\left(\frac{1}{1+i}\right)}^{12t}}$

Formula- $2$:

  $M=\frac{Pi{\left(1+i\right)}^{12t}}{{\left(1+i\right)}^{12t}-1}$

Formula used:

  $M=\frac{Pi}{1-{\left(\frac{1}{1+i}\right)}^{12t}}$

Proof:

Given formula-

  $M=\frac{Pi}{1-{\left(\frac{1}{1+i}\right)}^{12t}}$

Distribute the exponent,

  $M=\frac{Pi}{1-\frac{{\left(1\right)}^{12t}}{{\left(1+i\right)}^{12t}}}$

Apply; ${\left(1\right)}^n=1$,

  $M=\frac{Pi}{1-\frac{1}{{\left(1+i\right)}^{12t}}}$

Now, simplify;

  $\begin{array}{l}M=\frac{Pi}{1-\frac{1}{{\left(1+i\right)}^{12t}}}\\ M=\frac{Pi{\left(1+i\right)}^{12t}}{{\left(1+i\right)}^{12t}-1}\end{array}$

b.

To calculate: Find the monthly payment.

The monthly payment is $\$364.02$ .

Given information:

Borrow money $\left(P\right)$ : $\$15,500$

Interest rate: $0.5\%$

Repay the loan over time $\left(t\right)$ : $4$ year

Formula used:

  $M=\frac{Pi{\left(1+i\right)}^{12t}}{{\left(1+i\right)}^{12t}-1}$

Calculation:

Given that-

  $P=15500$ ;

  $t=4$ ;

  $i=0.005$ ;

Now, plug the all values in formula-

  $\begin{array}{l}M=\frac{Pi{\left(1+i\right)}^{12t}}{{\left(1+i\right)}^{12t}-1}\\ \\ M=\frac{\left(15500\right)\left(0.005\right){\left(1+0.005\right)}^{12\times 4}}{{\left(1+0.005\right)}^{12\times 4}-1}\\ \\ M=\$364.02\end{array}$

Conclusion:

  $M=\$364.02$