1. ## Geometric Series

A rubber ball is dropped from height 10 meters. If it rebounds aprox 1/2 the distance after each fall, use geometric series to aprox the total distance that the ball travels before coming to rest.

2. Hello, ggcoralysp!

A rubber ball is dropped from height 10 meters.
If it rebounds aprox 1/2 the distance after each fall, use geometric series
to approximate the total distance that the ball travels before coming to rest.
Let's baby-talk our way through this . . .

The ball falls 10 meters: .$\displaystyle 10$

It bounces up 5 meters and falls 5 meters: .$\displaystyle 2(5)$

It bounces up $\displaystyle \tfrac{5}{2}$ meters and falls $\displaystyle \tfrac{5}{2}$ meters: .$\displaystyle 2\left(\tfrac{5}{2}\right)$

It bounces up $\displaystyle \tfrac{5}{2}$ meters and falls $\displaystyle \tfrac{5}{2}$ meters: .$\displaystyle 2\left(\tfrac{5}{2}\right)$

It bounces up $\displaystyle \tfrac{5}{4}$ meters and falls $\displaystyle \tfrac{5}{4}$ meters: .$\displaystyle 2\left(\tfrac{5}{4}\right)$

. . and so on . . .

The total distance is: .$\displaystyle D \;=\;10 + 2(5) + 2\left(\tfrac{5}{2}\right) + 2\left(\tfrac{5}{4}\right) + 2\left(\tfrac{5}{8}\right) + \hdots$

$\displaystyle \text{We have: }\;D \;=\;10 + 10\underbrace{\bigg[1 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \hdots\bigg]}_{\text{geometric series}}$ .[1]

The geometric series has: .first term $\displaystyle a = 1$, common ratio $\displaystyle r = \tfrac{1}{2}$
. . Its sum is: .$\displaystyle \frac{1}{1-\frac{1}{2}} \:=\:2$

Substitute into [1]: .$\displaystyle D \;=\;10 + 10(2) \;=\;30$ meters.