Answer
3 m
Work Step by Step
Let $a_{n}$ be the height the ball after the nth bounce.
Given: $a_{0}=1$ and $a_{n}=\displaystyle \frac{1}{2}a_{n-1}$.
$a_{1}=\displaystyle \frac{1}{2}, $
$a_{2}=\displaystyle \frac{1}{2}(\frac{1}{2})=(\frac{1}{2})^{2}$
$a_{3}=\displaystyle \frac{1}{2}(\frac{1}{2})^{2}=(\frac{1}{2})^{3}$
$...$
$a_{n}=(\displaystyle \frac{1}{2})^{n}$
The total distance traveled =
$a_{0}+$
$+2a_{1}\quad$ ($a_{1}$ up, $a_{1}$ down),
$+2a_{2}\quad$ ($a_{2}$ up, $a_{2}$ down),
$+2a_{3}\quad$ ($a_{3}$ up, $a_{3}$ down),
$+...$
$=1+2(\displaystyle \frac{1}{2})+2(\frac{1}{2})^{2}+2(\frac{1}{2})^{3}+2(\frac{1}{2})^{4}+\cdots$
$=1+(1+\displaystyle \frac{1}{2}+(\frac{1}{2})^{2}+(\frac{1}{2})^{3}+\cdots)$
$=1+\displaystyle \sum_{k=1}^{\infty}(\frac{1}{2})^{k-1}\qquad $... geometric series,
... a=1, $r=\displaystyle \frac{1}{2} < 1$. so it converges to $\displaystyle \frac{a}{1-r}$ ...
$=1+\displaystyle \frac{1}{1-\frac{1}{2}}$
$=1+\displaystyle \frac{1}{\frac{1}{2}}=1+2$
$=3$
