Let's express this as a sum:
sum_(k=1)^oo 500(1.12)^-k
Since 1.12=112/100=28/25, this is equivalent to:
sum_(k=1)^oo 500(28/25)^-k
Using the fact that (a/b)^-c=(1/(a/b))^c=(b/a)^c, we have:
sum_(k=1)^oo 500(25/28)^k
Also, we can pull the 500 out of the summation sign, like this:
500sum_(k=1)^oo (25/28)^k
Alright, now what is this? Well, sum_(k=1)^oo(25/28)^k is what's known as a geometric series. Geometric series involve an exponent, which is exactly what we have here. The awesome thing about geometric series like this one is that they sum up to r/(1-r), where r is the common ratio; i.e. the number that's raised to the exponent. In this case, r is 25/28, because 25/28 is what's raised to the exponent. (Side note: r has to be between -1 and 1, or else the series doesn't add up to anything.)
Therefore, the sum of this series is:
(25/28)/(1-25/28)
=(25/28)/(3/28)
=25/28*28/3=25/3
We've just discovered that sum_(k=1)^oo (25/28)^k=25/3, so the only thing that's left is to multiply it by 500:
500sum_(k=1)^oo (25/28)^k
=500*25/3
=12500/3~~4166.667
You can find out more about geometric series here (I encourage you to watch the whole series Khan Academy has on geometric series).
