Question 1179300: A rubber ball is dropped from a height of 100 feet, and at each bounce it rebounds one-half of the height from which it last fell. What distance has the ball traveled up to the instant it hits the ground for the eight time? Solve using a geometric sequence.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! First time, it drops 100 feet
Then it rises 50 and drops 50 for second time and so forth.
100+(100)(1/2)+100(1/2) for the second, and we only care about the distance traveled, not the direction.
100+100(1/2)+100(1/2)+100(1/2)^2+100(1/2)^2 +...+100(1/2)^7+100(1/2)^7.
Leave out the first term for a moment.
the other terms are 200(1/2)+200(1/2)^2+200(1/2)^3+200(1/2)^4+200(1/2)^5+200(1/2)^6+200(1/2)^7
The geometric sequence for sum of (1/2)^n for n=1-7 is a*(1-r^n)/(1-r); a, the first term, is 100
This is here 100(1-(1/128))/(1/2)
This is 100(63/64)*2=400*63/64=198.4375 feet
Now bring in the other term 100 for the first drop, and the answer is 298.4375 feet.
Check this
down 100
up 50 down 50 (100)
up 25 down 25 (50)
up 12.5 down 12.5 (25)
then 12.5
then 6.25
then 3.125
then 1.5625
That sum is 298.4375 feet
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