problem: A ball is dropped from $405$ meters and rebounds two-thirds the distance it falls each time it bounces. How many meters will the ball have traveled when it hits the ground the fourth time?
solution: We can split the ball's motion into two parts: when it goes down, and when it goes up. By adding these two parts together separately, we get two geometric series.
  
  We will first calculate when the total distance that the ball falls. Initially, it falls $405$ meters. The next time, it will have rebounded $405(2/3)$ meters, so it will also fall that much. The next time, it will have rebounded $405(2/3)(2/3)$ meters, and so on. So, we have a finite geometric series with first term $405$ with common ratio $2/3.$ Since the ball falls four times before hitting the ground the fourth time, there are four terms in this series. The total distance that the ball falls is therefore $$\frac{405\left(1-\left(\frac23\right)^4\right)}{1-\frac23} = 975.$$Now, we calculate the total distance that the ball rises. Initially, the ball rises $405(2/3)$ meters. The next time, it rises $405(2/3)(2/3)$ meters, and so on. This time, our geometric series has a first term of $405(2/3),$ common ratio $2/3,$ and three terms. Thus, the ball rises a total of $$\frac{405\cdot\frac23\left(1-\left(\frac23\right)^3\right)}{1-\frac23} = 570.$$Adding together these two values, we find that the ball has traveled a total of $975 + 570 = \boxed{1545}$ meters.