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I'm having a bit of trouble determining what I assume to be an infinite geometric sum for the following question in part ii.)

I've posted my working below, but I'm not sure how to get the geometric sum in the form the question requires.

Working for Geometric Sum Problem:

enter image description here

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  • $\begingroup$ Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. $\endgroup$ Aug 27, 2022 at 8:07
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    $\begingroup$ Remember that whatever goes up must come down $\endgroup$
    – Paul
    Aug 27, 2022 at 9:17
  • $\begingroup$ Would it then be H + 2(q/100) + 2(q^2/100^2) + ... etc.? $\endgroup$ Aug 27, 2022 at 9:26
  • $\begingroup$ @GeorgeOrwell Yes $\endgroup$
    – FShrike
    Aug 27, 2022 at 9:29
  • $\begingroup$ Don't forget the H factor $H(1 +\frac{2q}{100}+ ...)$ and $H=2H-H$ is handy $\endgroup$
    – Paul
    Aug 27, 2022 at 9:35

1 Answer 1

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You're forgettingthat the ball first falls with height H, then rises to height $\frac{qH}{100}$, falls with height $\frac{qH}{100}$ and so on.

Therefore, the correct distance must be :

H + 2(S)

where S is the sum of heights starting from $\frac{qH}{100}$.

$S = \frac{\frac{qH}{100}}{1 - \frac{q}{100}}$

Simplifying, $S = \frac{qH}{100 - q}$

Going back to the original problem,

$H + 2(\frac{qH}{100 - q})$

$\frac{100H - qH + 2qH}{100 - q}$

$\frac{100H + qH}{100 - q}$

$H(\frac{100 + q}{100 - q})$

which is the required answer.

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  • $\begingroup$ Ah, right. I had to treat the initial height seperate from the remaining geometric series there. Thank you! $\endgroup$ Aug 27, 2022 at 10:21

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