Question 109830
<pre><font face = "book antiqua" size = 5><b>
{{{drawing(200,400,0,20,-1,20,
line(0,16,1,0), line(1,0,2,8),line(2,8,3,0),line(3,0,4,4),
line(4,4,5,0),line(5,0,6,2),line(6,2,7,0),line(7,0,8,1),
line(8,1,9,0),line(9,0,10,.5),line(10,.5,11,0),
line(11,0,12,.25),line(12,.25,13,0), line(13,0,14,1/8),
line(14,1/8,15,0),line(15,0,16,1/16), line(16,1/16,17,0)

line(17,0,18,1/32),line(18,1/32,19,0)

  )}}}    
From the picture, you can see that the ball falls 10 times 
but it only rises 9 times.

The "falls" form this geometric series with 10 terms, with
first term 16, common ratio 1/2:

16+8+4+2+1+1/2+1/4+1/8+1/16+1/32

The "rises" form this geometric series with 9 terms, with
first term 8, common ratio 1/2

8+4+2+1+1/2+1/4+1/8+1/16+1/32

The only difference in the two is the first "fall" of 16 ft. 
Let's find the sum of the geometric series of "rises" and then 
we can find the sum of the geometric series of "falls" by adding 
the first 16 foot "fall" to it:

The formula for the sum of a geometric series is

S<sub>n</sub> = a<sub>1</sub>(1-r<sup>n</sup>)/(1-r)

For the "rises", a<sub>1</sub> = 8, r = 1/2, n = 9

S<sub>9</sub> = 8[1-(1/2)<sup>9</sup>]/(1-1/2)

S<sub>9</sub> = 8[1-1/512]/(1/2)

S<sub>9</sub> = 8[511/512]×(2/1)

S<sub>9</sub> = (8×511×2)/(512)

S<sub>9</sub> = 511/32 feet = sum of "rises" only.

The sum of the "falls" accounts for 16 feet more
than the "rises" or

Sum of "falls" = 511/32 + 16 = 511/32 + 512/32 = 1023/32 feet

So, "sum of rises" + "sum of falls" = 511/32 + 1023/32 = 

1534/32 = 767/16 feet = 47.9375 feet the ball travels.

Edwin</pre>