1. ## Sigma question...

Hey guys!

I just wanted to say that I really appreciate this forum. I'm doing some power studying for a college algebra test (I'm in high school) so I appreciate the help with the "power studying" I'm doing.

Take a look at this sigma-related problem...

A ball is dropped from a height of h feet and repeatedly bounces off the floor. After each bounce, the ball reaches a height that is $\displaystyle \frac2 3$ of the height from which is previously fell. For example, after the first bounce, the ball reaches a height of $\displaystyle \frac2 3$h feet. Which of the following represents the total number of feet the ball travels between the first and the sixth bounce?
Answer: $\displaystyle \sum_{i=1}^5$ (2h)$\displaystyle (\frac2 3)^i$

I understand how sigma works and stuff, but why 2h$\displaystyle (\frac2 3)^i$? Why the "2" in 2h? I don't see anywhere in the actual problem where is calls to double something? Thanks... I'm sure it's something simple I'm forgetting...

2. Originally Posted by theimcq
Hey guys!

I just wanted to say that I really appreciate this forum. I'm doing some power studying for a college algebra test (I'm in high school) so I appreciate the help with the "power studying" I'm doing.

Take a look at this sigma-related problem...

Answer: 5 $\displaystyle \Sigma$ i=1 (2h)$\displaystyle (\frac2 3)^i$

I understand how sigma works and stuff, but why 2h$\displaystyle (\frac2 3)^i$? Why the "2" in 2h? I don't see anywhere in the actual problem where is calls to double something? Thanks... I'm sure it's something simple I'm forgetting...
Thats because the ball has to fall the same distance it goes up and thus covers that distance twice...

3. Wow... I really am not thinking straight. ROFL! Thanks for pointing this very, very elementary fact out.

4. Hello,

Originally Posted by theimcq
Hey guys!

I just wanted to say that I really appreciate this forum. I'm doing some power studying for a college algebra test (I'm in high school) so I appreciate the help with the "power studying" I'm doing.

Take a look at this sigma-related problem...

Answer: 5 $\displaystyle \Sigma$ i=1 (2h)$\displaystyle (\frac2 3)^i$

I understand how sigma works and stuff, but why 2h$\displaystyle (\frac2 3)^i$? Why the "2" in 2h? I don't see anywhere in the actual problem where is calls to double something? Thanks... I'm sure it's something simple I'm forgetting...
1/ Sigma is equivalent to "sum for i going from 1 to 5 of the following expression"
for example (a simple one) :

$\displaystyle \sum_{n=1}^4 (2n)=(2*1)+(2*2)+(2*3)+(2*4)$

It's the sum of the expressions with all the possible values of n, that is to say 1 to 4. I don't know if it's clear enough...

2/
Hehe, I'll try to do it the way Soroban does

$\displaystyle \underbrace{\downarrow}_{\text{1st bounce}} \ \underbrace{\color{red}\uparrow \ \downarrow}_{h(2/3)} \ \underbrace{\color{blue}\uparrow \ \downarrow}_{h(2/3)^2} \ \dots$

Edit : too slow

5. Yeah... perhaps I need to go over some simple Physics again hahahah! Bounce = up and down. LOL! Your post is as equally helpful! Thanks again!