Let's express this as a sum:
∞∑k=1500(1.12)−k
Since 1.12=112100=2825, this is equivalent to:
∞∑k=1500(2825)−k
Using the fact that (ab)−c=(1ab)c=(ba)c, we have:
∞∑k=1500(2528)k
Also, we can pull the 500 out of the summation sign, like this:
500∞∑k=1(2528)k
Alright, now what is this? Well, ∞∑k=1(2528)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 r1−r, where r is the common ratio; i.e. the number that's raised to the exponent. In this case, r is 2528, because 2528 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:
25281−2528
=2528328
=2528⋅283=253
We've just discovered that ∞∑k=1(2528)k=253, so the only thing that's left is to multiply it by 500:
500∞∑k=1(2528)k
=500⋅253
=125003≈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).
