# 13.5 Extra credit

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An extra problem on compound interest to be used for extra credit. This module is part of the Teacher's Guide.

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

A bank gives $i%$ interest, compounded annually. (For instance, if $i=6$ , that means 6% interest.) You put $A$ dollars in the bank every year for $n$ years . At the end of that time, how much money do you have?

(The fine print: Let’s say you make your deposit on January 1 every year, and then you check your account on December 31 of the last year. So if $n=1$ , you put money in exactly once, and it grows for exactly one year.)

The money you put in the very last year receives interest exactly once.“Receiving interest”in a year always means being multiplied by $\left(1+\frac{i}{\text{100}}\right)$ . (For instance, if you make 6% interest, your money multiplies by 1.06.) So the $A$ dollars that you put in the last year is worth, in the end, $A\left(1+\frac{i}{\text{100}}\right)$ .

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.

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

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

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

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

Example: If you invest $5,000 per year at 6% interest for 30 years, you end up with: $\frac{\text{100}\left(\text{5000}\right)}{6}\left[1.0631–1.06\right]=419,008.39$ Not bad for a total investment of$150,000!

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