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# HELP ON A NEW (probably never been heard) COURSEWORK!! watch

1. Hi, im doing a statistics coursework for my gcse, but im in the middle of the extension, and im stuck. below is what i am asked to do:

Jo now is asked to construct FREE-STANDING tins in the middle of the floor.

Investigate possible free-standing tin stacks with, for example, square and other polygonal bases, making the rules for such construction.

For each questions when there is a formula discovered, explain what it is and how it was found.

so i was wondering if anyone could help, because i am so confused!!!
2. Is your problem an extension of the one at http://www.uk-learning.net/showthread.php?t=54845 ?
3. the one you are talking about is an earlier part of my coursework, now im on the extension
4. Suppose the bottom layer is a k-sided polygon with side length n.

Suppose that each layer above the bottom one has side length one less that the layer below it.

Suppose that there are m layers in total. (We must have m <= n. If m = n then we have a pyramid. If m < n then we have a pyramid with its top cut off.)

Count the dots:

* layer 1 has 1 + (1/2)kn(n - 1) dots,
* layer 2 has 1 + (1/2)k(n - 1)(n - 2) dots,
* layer 3 has 1 + (1/2)k(n - 2)(n - 3) dots,
* etc,
* layer m has 1 + (1/2)k(n - m + 1)(n - m) dots.

These numbers add up to m + (1/2)k*[S(n) - S(n - m)], where

S(n) = 2*1 + 3*2 + 4*3 + 5*4 + ... + n*(n - 1).

We can get a formula for S(n) by the usual method of calculating differences. The answer is

S(n) = (1/3)(n - 1)n(n + 1).

Putting that into the red formula, it follows that the total number of dots is

m + (1/6)k*[(n - 1)n(n + 1) - (n - m - 1)(n - m)(n - m + 1)].

As a check, consider the example where k = 4, n = 5 and m = 3:

* layer 1 has 1 + (1/2)4*5*4 = 41 dots,
* layer 2 has 1 + (1/2)4*4*3 = 25 dots,
* layer 3 has 1 + (1/2)4*3*2 = 13 dots.

These numbers add up to 79. The formula gives 3 + (2/3)*[4*5*6 - 1*2*3] = 3 + (2/3)*[120 - 6] = 3 + (2/3)*114 = 3 + 2*38 = 79.
5. It seems that they only want you to look at whole pyramids, not pyramids with the top cut off. For such stacks, we can replace m by n in the formula above to get

n + (1/6)k(n - 1)n(n + 1).

This formula covers a lot of cases. For example, in the attached picture we have

k = 6 (hexagonal base),
n = 5 (side-length 5 on the base).

Layer 1 has 61 cans.
Layer 2 has 37 cans.
Layer 3 has 19 cans.
Layer 4 has 7 cans.
Layer 5 has 1 can.
In total there are 125 cans.

The formula gives 5 + (1/6)6*4*5*6 = 5 + 4*5*6 = 125.
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