# Infinite Series.

• Sep 21st 2011, 08:20 PM
Cooper
Infinite Series.
I'm becoming increasingly irritated with a few questions given to me from my teacher about this subject, and would appreciate some help. They all seem extremely simple, but for some reason my answers, which seems perfectly logical keep coming out wrong when compared to the answer sheet.

1) A Ball is dropped from a height of 2m to a floor, on each bounce the ball rises to 50% of the height of which it fell, calculate the total distance the ball travels before coming to a rest.

Now this one is pretty mundane, I mean, 1m/2, = .5 or 1/2, so the ratio is a given.
I put it into the formula, (a/1-r) and get 4, which is apparently wrong. The correct answer is 6m. This one is frustrating me the most, because I'm normally very good at algebra, and it seems that every time I try it I get the wrong answer. So I mean, there has to be something wrong with what I'm doing in general, since I've been using the formula right... (2/-0.5)

2) My second problem goes as follows:
The midpoints of a square with sides 1, long are joined to form another square. Then the midpoints of the sides of the second square are joined to form a third square, and this process repeats infinitely. Find the total length of the segments.

Same problem, I use the formula, get 8. The correct answer is (8 + 4*root2).

I would really appreciate some help.
Thanks to anyone who responds.
• Sep 21st 2011, 08:47 PM
alexmahone
Re: Infinite Series.
Quote:

Originally Posted by Cooper
I'm becoming increasingly irritated with a few questions given to me from my teacher about this subject, and would appreciate some help. They all seem extremely simple, but for some reason my answers, which seems perfectly logical keep coming out wrong when compared to the answer sheet.

1) A Ball is dropped from a height of 2m to a floor, on each bounce the ball rises to 50% of the height of which it fell, calculate the total distance the ball travels before coming to a rest.

Now this one is pretty mundane, I mean, 1m/2, = .5 or 1/2, so the ratio is a given.
I put it into the formula, (a/1-r) and get 4, which is apparently wrong. The correct answer is 6m. This one is frustrating me the most, because I'm normally very good at algebra, and it seems that every time I try it I get the wrong answer. So I mean, there has to be something wrong with what I'm doing in general, since I've been using the formula right... (2/-0.5)

2) My second problem goes as follows:
The midpoints of a square with sides 1, long are joined to form another square. Then the midpoints of the sides of the second square are joined to form a third square, and this process repeats infinitely. Find the total length of the segments.

Same problem, I use the formula, get 8. The correct answer is (8 + 4*root2).

I would really appreciate some help.
Thanks to anyone who responds.

The ball drops 2 m, rises to 1 m, drops 1 m, rises to 1/2 m, drops 1/2 m and so on.

1) $h=2+1+1+\frac{1}{2}+\frac{1}{2}...$

$=2+2*\frac{1}{1-\frac{1}{2}}$

$=6$

2) Length of the side of the second square = $\frac{1}{\sqrt{2}}$

$S=4(1+\frac{1}{\sqrt{2}}+\frac{1}{2}+...)$

$=4*\frac{1}{1-\frac{1}{\sqrt{2}}}$

$=4*\frac{\sqrt{2}}{\sqrt{2}-1}$

$=4\sqrt{2}(\sqrt{2}+1)$

$=8+4\sqrt{2}$
• Sep 21st 2011, 09:08 PM
Cooper
Re: Infinite Series.
Quote:

Originally Posted by alexmahone
2) Length of the side of the second square = $\frac{1}{\sqrt{2}}$

$S=4(1+\frac{1}{\sqrt{2}}+\frac{1}{2}+...)$

$=4*\frac{1}{1-\frac{1}{\sqrt{2}}}$

$=4*\frac{\sqrt{2}}{\sqrt{2}-1}$

$=4\sqrt{2}(\sqrt{2}+1)$

$=8+4\sqrt{2}$

Thank you, but may I ask how you came to the conclusion that the side length of the second square = 1/root2? I'm sure it's something simple that I'm overlooking, but I would appreciate some clarification.
• Sep 21st 2011, 09:12 PM
alexmahone
Re: Infinite Series.
Quote:

Originally Posted by Cooper
Thank you, but may I ask how you came to the conclusion that the side length of the second square = 1/root2? I'm sure it's something simple that I'm overlooking, but I would appreciate some clarification.

Refer to the attached figure.