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

1 Answer
Jul 22, 2016

The answer is either 40/9 or 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(2/3)^2 = 10(4/9) = 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(2/3)^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 ar^n

In this case, our series has a = 10 and r = 2/3.

We will also note that:
sum_(n=0)^infty ar^n = asum_(n=0)^infty r^n

So we can simply compute the sum of a geometric series (2/3)^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 = 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 + 2/3 + 4/9 + 8/27 + ...

We can see that:
sum_(n=2)^infty 10(2/3)^n = 10sum_(n=2)^infty (2/3)^n = 10[sum_(n=0)^infty (2/3)^n - (1 + 2/3)]
=10[1/(1-(2/3)) - (1 + 2/3)]
= 10[3 - 5/3] = 10[9/3 - 5/3] = 40/3