I now how to workout this problem, but I make a few small errors along the way and I need some sort of checking or help.

Problem.

A ball is drooped from a height of 10 feet (Assume no air resistance), each bounce is $\displaystyle (\frac{3}{4}) $of the height of the bounce before. Thus, after the balls hit the floor for the first time, the ball rises to a height of $\displaystyle 10(\frac{3}{4})$, after the second time, $\displaystyle 10(\frac{3}{4})^2$.

A. Find an expression for the height which the ball rises

Problem. after it hits the floor for the nth time.

$\displaystyle \sum_{n=0}^{\infty} 10\cdot\frac{3}{4}^n$

Thus, $\displaystyle H_0 = 10$, $\displaystyle H_1 = 10\frac{3}{4}$, $\displaystyle H_2 = 10(\frac{3}{4})^2$ $\displaystyle H_n = 10(\frac{3}{4})^n$

B. Find an expression for the total vertical distance the ball has travled for the first, second, third, and fourth times.

$\displaystyle D_n=H_0+2H_1+2H_3.......+2H_{n-1}$

$\displaystyle D_1=10+2\cdot10(\frac{3}{4})$

$\displaystyle D_2=10+2\cdot10(\frac{3}{4})+2\cdot10(\frac{3}{4}) ^2$

$\displaystyle D_3=10+2\cdot10(\frac{3}{4})+2\cdot10(\frac{3}{4}) ^2+2\cdot10(\frac{3}{4})^3$

$\displaystyle D_3=10+2\cdot10(\frac{3}{4})+2\cdot10(\frac{3}{4}) ^2+2\cdot10(\frac{3}{4})^3+2\cdot10(\frac{3}{4})^4$

D. Find the closed form expression for the total distance traveled for the nth time, this the part where I get my confused in properly making a simplify formula and simplifying it