# Question #f8e6c

Aug 24, 2016

Express it as a geometric series to find the sum is $\frac{12500}{3}$.

#### Explanation:

Let's express this as a sum:
${\sum}_{k = 1}^{\infty} 500 {\left(1.12\right)}^{-} k$

Since $1.12 = \frac{112}{100} = \frac{28}{25}$, this is equivalent to:
${\sum}_{k = 1}^{\infty} 500 {\left(\frac{28}{25}\right)}^{-} k$

Using the fact that ${\left(\frac{a}{b}\right)}^{-} c = {\left(\frac{1}{\frac{a}{b}}\right)}^{c} = {\left(\frac{b}{a}\right)}^{c}$, we have:
${\sum}_{k = 1}^{\infty} 500 {\left(\frac{25}{28}\right)}^{k}$

Also, we can pull the $500$ out of the summation sign, like this:
$500 {\sum}_{k = 1}^{\infty} {\left(\frac{25}{28}\right)}^{k}$

Alright, now what is this? Well, ${\sum}_{k = 1}^{\infty} {\left(\frac{25}{28}\right)}^{k}$ 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 $\frac{r}{1 - r}$, where $r$ is the common ratio; i.e. the number that's raised to the exponent. In this case, $r$ is $\frac{25}{28}$, because $\frac{25}{28}$ 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:
$\frac{\frac{25}{28}}{1 - \frac{25}{28}}$

$= \frac{\frac{25}{28}}{\frac{3}{28}}$

$= \frac{25}{28} \cdot \frac{28}{3} = \frac{25}{3}$

We've just discovered that ${\sum}_{k = 1}^{\infty} {\left(\frac{25}{28}\right)}^{k} = \frac{25}{3}$, so the only thing that's left is to multiply it by $500$:
$500 {\sum}_{k = 1}^{\infty} {\left(\frac{25}{28}\right)}^{k}$
$= 500 \cdot \frac{25}{3}$
$= \frac{12500}{3} \approx 4166.667$

You can find out more about geometric series here (I encourage you to watch the whole series Khan Academy has on geometric series).