# How do you find the sum of the infinite geometric series 10(2/3)^n when n=2?

Jul 22, 2016

The answer is either $\frac{40}{9}$ or $\frac{40}{3}$ depending on what was meant by the question.

#### Explanation:

Well if $n = 2$ then there isn't a sum, the answer is just:
$10 {\left(\frac{2}{3}\right)}^{2} = 10 \left(\frac{4}{9}\right) = \frac{40}{9}$

But perhaps the question was meant to ask that the infinite sum be taken starting at $n = 2$ such that the equation is:

${\sum}_{n = 2}^{\infty} 10 {\left(\frac{2}{3}\right)}^{n}$

In this case, we would compute it by first noting that any geometric series can be seen as being of the form:
${\sum}_{n = 0}^{\infty} a {r}^{n}$

In this case, our series has $a = 10$ and $r = \frac{2}{3}$.

We will also note that:
${\sum}_{n = 0}^{\infty} a {r}^{n} = a {\sum}_{n = 0}^{\infty} {r}^{n}$

So we can simply compute the sum of a geometric series ${\left(\frac{2}{3}\right)}^{n}$ and then multiply that sum by $10$ to arrive at our result. This makes things easier.

We also have the equation:
${\sum}_{n = 0}^{\infty} {r}^{n} = \frac{1}{1 - r}$

This allows us to compute the sum of the series starting from $n = 0$. But we want to compute it from $n = 2$. In order to do this, we will simply subtract the $n = 0$ and $n = 1$ terms from the full sum. Writing the first several terms of the sum out we can see that it looks like:
$1 + \frac{2}{3} + \frac{4}{9} + \frac{8}{27} + \ldots$

We can see that:
${\sum}_{n = 2}^{\infty} 10 {\left(\frac{2}{3}\right)}^{n} = 10 {\sum}_{n = 2}^{\infty} {\left(\frac{2}{3}\right)}^{n} = 10 \left[{\sum}_{n = 0}^{\infty} {\left(\frac{2}{3}\right)}^{n} - \left(1 + \frac{2}{3}\right)\right]$
$= 10 \left[\frac{1}{1 - \left(\frac{2}{3}\right)} - \left(1 + \frac{2}{3}\right)\right]$
$= 10 \left[3 - \frac{5}{3}\right] = 10 \left[\frac{9}{3} - \frac{5}{3}\right] = \frac{40}{3}$