How do you apply finite geometric series in order to determine the distance a bouncing ball travels up and down at the 10th bounce?

Question

My question is very similar to this one, except I would like to determine the distance traveled immediately after the 10th bounce.

Assume the ball is let go from 1 meter above the ground and each successive bounce is 2/3 the height of the previous bounce.

So the series is 1 + 2 * (2/3 + 4/9 + 8/27 + ...) = 1 + 2 ( 2 ) = 5 (by applying infinite geometric)

So obviously the final answer for the distance the ball travels by the 10th bounce must be less than 5.

At the 10th bounce the distance the ball travels should be:

1 + $2\sum_{k=0}^{9}(\frac{2}{3})^{k+1} = 1 + 4(1-(\frac{2}{3})^{10}) = 5 - 4(\frac{2}{3})^{10}\approx 4.93$ meters

However the solution to this problem says after the 10th bounce the ball has traveled $6 (\frac{2}{3})^{10} \approx .104 $ meters.

What am I doing wrong?

You seem to me to be calculating how far the ball travels until the end of the 10th bounce, and you're asked for how far it travels after the 10th bounce. I get the answer stated, you need to add up $\sum_{10}^{\infty}2\cdot \frac{2}{3}^n$ (or maybe start at $n=11$ — ancient mathematician, Feb 02 '21 at 20:21

score 1 · Accepted Answer · answered Feb 03 '21 at 07:52

I think you have misunderstood the question and calculated how far the ball travels before rather than after the 10th bounce.

After the 10th bounce the distance travelled is $\sum_{k=10}^{\infty} 2\cdot(\frac{2}{3})^k$ which is the required $6\cdot (\frac{2}{3})^{10}$.

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