Series: Every time the ball bounces, it rebounds to a height

[flakine]

A certain ball has the property that each time it falls from a height b onto a hard, level surface, it rebounds to a height nb, where 0<n<1.

Suppose that the ball is dropped from an initial height of B meters.

Assuming that the ball continues to bounce indefinitely, find the total distance that it travels in meters.

I have: the ball falls from height b, then rebounds to height nb, then falls nb, then rebounds to n(nb), then falls to n(nb)......

b+2nb+2n^2+2n^3......

Could someone help me complete this problem?

 

[stapel]

Hint:

. . . . .b + 2bn + 2bn² + 2bn³ + ...

. . . . .b + 2b[n + n² + n³ + ...]

Can you find "n + n² + n³ + ..."...? Then multiply by "2b", and tack on the initial term "b"...?

Eliz.

 

[skeeter]

where did the factor "b" go in your later terms?

you should have the series ...

b + 2nb + 2n²b + 2n³b + ... =
b + 2b(n + n² + n³ + ... ) =
b[1 + 2(n + n² + n³ + ... )]

now ... what kind of series is (n + n² + n³ + ... ), and what is its sum?

 

[flakine]

n+n^2+n^3+n^4... is a geometric series, its sum is a(1-r^n)/1-r

 

[skeeter]

close ... it's an infinite geometric series with |n| < 1.

what's the sum of such a series?

 

[flakine]

a/1-r

 

[skeeter]

excellent! ... so, what distance does the ball travel in terms of b and n?

 

[flakine]

b[1+2(n/1-n)]
b[1+(2n/1-n)]
b[(1-n)/(1-n)+(2n/1-n)]
b[((1-n)+2n)/(1-n)]

Is this it?

 

[skeeter]

b[1+2(n/1-n)]
b[1+(2n/1-n)]
b[(1-n)/(1-n)+(2n/1-n)]
b[((1-n)+2n)/(1-n)]
one more step ... in the numerator, (1-n)+2n = ?

Is this it?

 

[flakine]

Got it! Thanks a lot!!!

b(1+n)/(1-n)