PDA

View Full Version : The Penny Problemo



Keikan
03-14-2004, 08:52 AM
Hi all! I got a homework problem that says "it would be more preferable to do it with other people" so could everyone here help me? please? My parents are busy nad they don't anything bout math.

The Penny Problem

Task 1: This is a image pattern. The first diagram has 1 circle. The second diagram has 5 circles. The third diagram has 12 circles. How many are in fourth? How many circles are in the next three diagrams?

http://keikan.250free.com/diagrams.gif
the diagram she supplied us with so I don't think it's +6 but i'm not sure.

Task 2: Is there a pattern to the number of circles you added each time? Make a list of all the patterns you notice.

Task 3: in math we want to generalize. Which means predict how the pattern continues. See if you can make a formula to the pattern.

Thats it but please show all your rough work. Thank You in advance.

Keikan
03-14-2004, 10:10 AM
my work so far

Image Resized
Image Resized
[img]http://keikan.250free.com/diagram1.gif' width='200' height='120' border='0' alt='click for full size view'> (http://keikan.250free.com/diagram1.gif)

Image Resized
[img]http://keikan.250free.com/task2.gif' width='200' height='120' border='0' alt='click for full size view'> ('http://keikan.250free.com/task2.gif')

bujub22
03-16-2004, 01:44 AM
:frusty: :frusty: me too

DirtyDan
03-16-2004, 02:00 AM
I realized most of this is rambling, so if you dont bother to look through it my equation was Tn = 1.5(n^2) - 0.5n :frusty:


Recursive:
T1 = 1
Tn = (Tn-1)-2+3n
T2 = 1-2+3(2) = 5
T3 = 5-2+3(3) = 12
T4 = 12-2+3(4)= 22
T5 = 22-2+3(5)= 35 hmmmmmmmm
Explicit:
Tn = T1 * r^(n-1) <----- i think thats right but i dunno how the hell to us it, so i made this up grrrrr :huh:

I found this by looking at the pattern
T2 = 2^2 + 2*.5
T3 = 3^2 + 3*1
T4 = 3^2 + 4*1.5.....
Tn = n^2+1.5n(n-1)
simplifies to.....
Tn = 1.5(n^2) - 0.5n :D
which works
T1 = 1.5(1*1)-0.5(1) = 1
T2 = 1.5(2*2)-0.5(2) = 5
T3 = 1.5(3*3)-0.5(3) = 12
T4 = 1.5(4*4)-0.5(4) = 22
So you could use that if you wanted to find the nth term, like how many would be in the 100th thingy would be
T100 = 1.5(100*100)-0.5(100) = 9950 :ph34r:
:angry: Im doing this stuff in math right now, i dunno what i did wrong with the explicit thing grrrr

Its past what your doing, but its one way you could get a formula for it im tired right now be back later :smilie4:

Keikan
03-17-2004, 04:33 AM
Wow&#39;s my math teacher would liek you but I don&#39;t think I should put a formula down since you don&#39;t have to make a formula you just need to find lots of patterns n&#39; such and i wouldn&#39;t be smart enough to think of that

DirtyDan
03-17-2004, 04:55 AM
What grade/class are you in keikan?

Keikan
03-17-2004, 06:57 AM
math 10 pure but I&#39;m in grade 9 but i&#39;m going to try cuz i got something good going

FlyingDutchman
03-20-2004, 02:41 PM
Ok Keikan,
Let&#39;s solve this in a graphical way, as the problem was presented in a graphical way.

At A you see the 4th image

At B I have changed the positions just a little bit.

http://www.bijlard.demon.nl/pics2/penny.gif

At C we split off the largest possible square. Amount = I^2

What rests is a triangle with a side always 1 less than the square&#39;s side

At D I added the purple row, and a triangle of the same size.
Now this amount is equal to our first square.

So what is the amount in the triangle?
((I^2)-I)/2


=========> amount in the total image = (I^2)+((I^2)-I)/2

I hope this helps.
Many problems can be solved in a graphical way. Try to understand what I did.