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This is an infinite geometric series question with r=.8 and the coefficient equal to 12. Since r<1, this series converges and the ball actually travels a finite distance.
$\displaystyle \Sigma_{k=0}^{k=\infty}r^k=\frac{1}{1-r}$ for |r|<1 that formula holds true, but in our case here, $\displaystyle \Sigma_{k=0}^{k=\infty}12(r)^k=\frac{12}{1-r}$ however I forgot that the ball bounces and falls so Soroban is correct
Hello, Sarah!
Let's baby-step our way through this . . .A ball is dropped from a height of 12 feet.
The ball hits the ground and rebounds 80% of the distance it falls each time.
In theory, how far will the ball travel before coming to rest?
It falls 12 feet.
$\displaystyle \text{Then it bounces up }80\%\text{ of }12\:=\:(0.8)12\text{ feet.}$
. . Then it falls that distance.
$\displaystyle \text{Then it bounces up }80\%\text{ of }(0.8)12 \:=\:(0.8)^212\text{ feet.}$
. . Then it falls that distance.
$\displaystyle \text{Then it bounces up }80\%\text{ of }(0.8)^212 \:=\:(0.8)^312\text{ feet.}$
. . Then it falls that distance.
And so on . . .
We can see that the total distance is:
. . $\displaystyle D \;=\;12 + 2(0.8)12 + 2(0.8)^212 + 2(0.8)^312 + 2(0.8)^412 + \hdots $
. . $\displaystyle D \;=\;12 + 2(0.8)12\underbrace{\bigg[1 + 0.8 + 0.8^2 + 0.8^3 + \hdots \bigg]}_{\text{geometric series}} $
The geometric series has first term $\displaystyle a = 1$, common ratio $\displaystyle r = 0.8$
. . Its sum is: .$\displaystyle \frac{1}{1-0.8} \:=\:\frac{1}{0.2} \:=\:5$
Therefore: .$\displaystyle D \;=\;12 + 2(0.8)12(5) \;=\;108\text{ feet}$