If a ball is dropped from a height of 10 meters, and on each bounce it reaches a maximal height of 0.75 times the previous height, then what is the total distance traveled? If we use h_i as the maximal height reached on bounce i, h_0 as the initial height, and $latex b $ as the factor of the maximal height achieved relative to the previous bounce, we have h_i = b h_{i-1}. Then, the total distance traveled is:

\begin{aligned}d &= h_0 + 2 h_1 + 2 h_2 + 2 h_3 + ...\\&= h_0 + 2 b h_0 + 2 b h_1 + 2 b h_2 + ...\\&= h_0 + 2 b h_0 + 2 b b h_0 + 2 b b b h_0 + ...\\&= h_0 + 2 b h_0 + 2 b^2 h_0 + 2 b^3 h_0 + ...\\&= -h_0 + 2 (h_0 + b h_0 + b^2 h_0 + b^3 h_0 + ...)\\&= -h_0 + 2 h_0 (1 + b + b^2 + b^3 + ...)\\&= -h_0 + 2 h_0 \sum_i \left(b^i\right)\end{aligned}

Using \sum_n \left(x^n\right) = \frac{1}{1-x} if 0 \le x < 1, we get:

d = -h_0 + 2 h_0 \frac{1}{1-b}

Now we plug our values of h_0 = 10 \text{ m} and b = 0.75 into this equation to find d = -10 + 20 \cdot \frac{1}{0.25} = 70 meters. So, even though our ball will bounce on for eternity, it will travel exactly 70 meters!