Distance traveled by a bouncing ball with exponentially diminishing rebounds This is kind of an odd question, but can somebody please tell me that I am crazy with the following question, I did the math, and what I am told to prove is simply wrong:
Question:
Show that a ball dropped from height of h feet and bounces in such a way that each bounce is $\frac34$ of the height of the bounce before travels a total distance of 7 h feet.
My Work:
$$\sum_{n=0}^{\infty} h \left(\frac34\right)^n = 4h$$
Obviously 4 h does not equal 7 h .  What does the community get?
I know that my calculations are correct, see Wolfram Alpha and it confirms my calculations, that only leaves my formula, or the teacher being incorrect...
Edit:
Thanks everyone for pointing out my flaw, it should be something like:
$$\sum_{n=0}^{\infty} -h + 2h \left(\frac34\right)^n = 7h$$
Thanks in advance for any help!
 A: Your computation does not give the total distance traveled, it only gives the distance it traveled downward.
The ball first falls $h$. Then it rises $\frac{3}{4}h$, and falls $\frac{3}{4}h$ again; then it rises $(\frac{3}{4})^2h$, and falls that much again. Etc.
So the total distance traveled by the ball is 
$$h + 2h\left(\frac{3}{4}\right) + 2h\left(\frac{3}{4}\right)^2 + \cdots = h + \sum_{n=0}^{\infty} \frac{3h}{2}\left(\frac{3}{4}\right)^{n-1}.$$
This gives, by the usual formula, a total distance of:
$$ h + \frac{\quad\frac{3h}{2}\quad}{1 - \frac{3}{4}} = h + \frac{\quad\frac{3h}{2}\quad}{\frac{1}{4}} = h + \frac{12h}{2} = 7h.$$
A: To get symmetry - a complete sawtooth pattern - suppose the ball is first thrown up from ground to height $h$. The total distance $x$ is that of the first tooth $= 2h$ plus the remaining sawtooth $= 3/4\ x$. Thus $x =\: 2h + 3/4\ x,\ $ so $x = 8h\:.\ $ Subtracting the intitial throw up leaves $7h.$
A: Before jumping to a formula, let us calculate a little.  The distance travelled until the first contact with the ground is $h$.  
The distance travelled between the first contact and the second is $(h)(2)(3/4)$
(up and then down).  The distance travelled from second contact to third is $(h)(2)(3/4)^2$, and so on.
So the total distance travelled is
$$h+(h)(2)\sum_{n=1}^{\infty}\left(\frac{3}{4}\right)^n$$
Finally, sum the infinite series.  That sum is $3$, giving a total of $7h$.
A: Hint: You are forgetting to count the distance traveled on the way up from the bounce. If the initial drop is from height $h$, the ball bounces up $\frac{3h}{4}$ and then drops from the same distance, and so on.
A: Note that when the ball bounces it goes both up and down. So from the second term onwards, you need to count each term twice. Therefore the answer is $2 \cdot 4h - h = 7h$ ($h$ is the first term, which is only counted once).
