13.5 Extra credit

An extra cool problem you may want to use as an extra credit or something

The previous year’s money receives interest twice, so it is worth $A{\left(1+\frac{i}{\text{100}}\right)}^2$ at the end. And so on, back to the first year, which is worth $A{\left(1+\frac{i}{\text{100}}\right)}^n$ (since that initial contribution has received interest $n$ times).

So we have a Geometric series:

$S=A\left(1+\frac{i}{\text{100}}\right)+A{\left(1+\frac{i}{\text{100}}\right)}^2+...+A{\left(1+\frac{i}{\text{100}}\right)}^n$

We resolve it using the standard trick for such series: multiply the equation by the common ratio, and then subtract the two equations.

$\left(1+\frac{i}{\text{100}}\right)S=A{\left(1+\frac{i}{\text{100}}\right)}^2+...+A{\left(1+\frac{i}{\text{100}}\right)}^n+A{\left(1+\frac{i}{\text{100}}\right)}^{n+1}$

$S=A\left(1+\frac{i}{\text{100}}\right)+A{\left(1+\frac{i}{\text{100}}\right)}^2+...+A{\left(1+\frac{i}{\text{100}}\right)}^n$

$\left(\frac{i}{\text{100}}\right)S=A{\left(1+\frac{i}{\text{100}}\right)}^{n+1}–A\left(1+\frac{i}{\text{100}}\right)$

$S=\frac{\text{100}A}{i}$ $\left[{\left(1+\frac{i}{\text{100}}\right)}^{n+1}-\left(1+\frac{i}{\text{100}}\right)\right]$