Associated with each rubber ball is a bounce coefficient b. When the ball is dropped from a height h, it bounces back to a height of b h. Suppose that the ball is dropped from an initial height h, and then is allowed to bounce forever.
Use the expansion of 1/(1 - x) in powers of x to help come up with a clean formula that measures, in terms of b and h, the total up-down distance the ball travels in all of its bouncing.

Question

Associated with each rubber ball is a bounce coefficient b.  When the ball is dropped from a height h, it bounces back to a height of b h.  Suppose that the ball is dropped from an initial height h, and then is allowed to bounce forever. 
Use the expansion of 1/(1 - x) in powers of x to help come up with a clean formula that measures, in terms of b and h, the total up-down distance the ball travels in all of its bouncing.

rogue · Answer

Oh, we did this last Friday! Hold on, let me get out my notes...

anonymous · Answer

1
 -   = 1/1- x )hb
 2

rogue · Answer

$$Height_{total} = H$$$$H = h + 2bh + 2b^2 h + 2b^3 h + ...$$$$H = h (1 + 2b + 2b^2 + 2b^3 + ...)$$$$H = h( 1 + 2b (1 + b + b^2 + b^3 +...))$$$$H = h \left[ 1 + 2b (\frac {1}{1-b}) \right] $$And that nicely becomes$$H = h \left[ \frac {1 + b}{1-b} \right]$$Done!

anonymous · Answer

Thank You!!!!!!

rogue · Answer

No problem, its a fun little geometric series challenge =)