<< Chapter < Page | Chapter >> Page > |
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 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.
$\left(1+\frac{i}{\text{100}}\right)S=A{\left(1+\frac{i}{\text{100}}\right)}^{2}+...+A{\left(1+\frac{i}{\text{100}}\right)}^{n}+A{\left(1+\frac{i}{\text{100}}\right)}^{n+1}$
$S=A\left(1+\frac{i}{\text{100}}\right)+A{\left(1+\frac{i}{\text{100}}\right)}^{2}+...+A{\left(1+\frac{i}{\text{100}}\right)}^{n}$
$\left(\frac{i}{\text{100}}\right)S=A{\left(1+\frac{i}{\text{100}}\right)}^{n+1}\u2013A\left(1+\frac{i}{\text{100}}\right)$
$S=\frac{\text{100}A}{i}$ $\left[{\left(1+\frac{i}{\text{100}}\right)}^{n+1}-\left(1+\frac{i}{\text{100}}\right)\right]$
Example: If you invest $5,000 per year at 6% interest for 30 years, you end up with:
$\frac{\text{100}(\text{5000})}{6}[1.0631\u20131.06]=\$\mathrm{419,008.39}$
Not bad for a total investment of $150,000!
Notification Switch
Would you like to follow the 'Advanced algebra ii: teacher's guide' conversation and receive update notifications?