I need help with Question 13, this is regarding the Loan Amortization Formula

Question

I have been working on 13, it is especially frustrating since I am so close to being done, but if there is anyone who could explain how to derive this, it would mean so much! I have retraced my work using Questions 4-6 and 11-12, but still no luck on getting the variable substitutions to work right!

Answer 1

First of all you have this equation:

$$B\cdot (1+i)^N=m\cdot \frac{(1+i)^N-1}{i} \qquad (*)$$

Explanation: Assume you borrow $B$ monetary units with an interest rate of $i\cdot 100$ percent per month. Then after N month the loan ($B$) has an value of $B\cdot (1+i)^N$. This is the future value of $B$. On the right hand side you have N monthly repayments of the loan. Every payment has a an amount of $m$. Every payment which is made at month $t$ is compounded so that it represents the future value as well. So you have an geometric series on the right hand side. Let $q=1+i$ then the right hand side is

$$m\cdot \sum_{x=0}^{N-1}q^x=m\cdot \left(q^0+q^1+q^2+\ldots +q^{N-1} \right)$$

The closed term for a partial sum of a geometric series is $$\sum_{x=0}^{N-1} q^x= \frac{q^N-1}{q-1}$$

This is what you have with the definition of q: $\frac{(1+i)^N-1}{(1+i)-1}=\frac{(1+i)^N-1}{i}$

Hopefully the equation $(*)$ ist clear now. Now we can start to solve the equation for $m$. Multiplying both sides by $i$.

$$B\cdot (1+i)^N\cdot i=m\cdot \left((1+i)^N-1\right) $$

Dividing the equation by $\left((1+i)^N-1\right)$

$$B\cdot \frac{(1+i)^N\cdot i}{(1+i)^N-1}=m $$

Next we expand the fraction on the right by $(1+i)^{-N}$. Every summand has to be multiplied by that factor. Here the exponent rules are helpful. For instance, $a^b\cdot a^{-b}=a^{b-b}=a^0=1$

$$B\cdot \frac{(1+i)^N\cdot (1+i)^{-N}\cdot i}{\underbrace{(1+i)^N\cdot (1+i)^{-N}}_{=1}-1\cdot (1+i)^{-N}}=m $$

$$\boxed{B\cdot \frac{ i}{1- (1+i)^{-N}}=m} $$