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# GCSE C/W - T TOTALS watch

1. Wondering if you can help me... Im currently doing the GCSE CW , T Totals, and am seeking help. A friend suggested i ask you...

If your not familiar with the CW, there are grids, and stairs. We must find the totals within the stairs. For instance on a 10 by 10 grid, and a 3 step square, the stair will cover 1 2 3 11 12 21, giving a total of 40ish.

Ive generated a forumlae for each stair 1-5, on any grid, shown below, but not a "master formula", for any stair size on any grid.

N is the 'base number', always on the bottom left of the stair, W, also called by many G, is the width of the grid

Step Size

1 1n + 0 + 0w
2 3n + 1 + 1w
3 6n + 4 + 4w
4 10n + 10 + 10w
5 15n + 20 + 20w

X ?n + _ + _w

i know that 1 3 6 10 and 15 are triganal numbers, thus the formula is ((n(n+1))/2) n + _ + _w

however i cannot work out _.

using differnce, T=an^3 + bn^2 + cn + d, i have worked out that 6a = 1, thus a = 1/6.... but do not know where to go from here.

Someone suggested it is 1/6n^3 - n/6, but i do not know how to get to here

All help much loved and apreciated, Sean
2. The terms in the sequence 1, 4, 10, 20, ... go up by the triangular numbers 1, 3, 6, 10, ...

4 is the sum of the first two triangular numbers: 1 + 3.
10 is the sum of the first three triangular numbers: 1 + 3 + 6.
20 is the sum of the first four triangular numbers: 1 + 3 + 6 + 10.
And so on.

Hence you need a formula for the sum of the first k triangular numbers. Google for "sum of triangular numbers".

3. thanks, but...

say i have a stair of 20010, can you tell me all the triangular numbers leading to 20010, and the total.

I commented that i know the triangular number formula, as i used to it work out ? x n, in the equation ?n + _ + _w.

If i remember correctly, its.. n(n+1) over 2
4. (Original post by sminggggg)
thanks, but...

say i have a stair of 20010, can you tell me all the triangular numbers leading to 20010, and the total.
Do you mean a staircase with 20010 steps? (Thinking big ...)

The formula will be

200210055n + _ + _w

where _ is the sum of the first 20009 triangular numbers.

--

If you Google as I suggested, you will find that the sum of the first k triangular numbers is (1/6)k(k + 1)(k + 2).

Since (1/6)*20009*20010*20011 = 1335334330165, the sum of the first 20009 triangular numbers is 1335334330165. So the formula is

200210055n + 1335334330165 + 1335334330165w
5. !! thank you johhny, you are my exalted god ....

only problem is now, howd i explain this on paper. I cant simply write...

"now i have the formula for working out the amount of "N" in the stair, i must work out the sum in each stair, using a formula..."
(1/6)k(k + 1)(k + 2) I found this by asking Johnny W"

So infact, the formula for ANY stair, on ANY size grid is infact.... cue drum roll...

T = (x(x-1) over 2) * n + (1/6)k(k + 1)(k + 2) + (1/6)k(k + 1)(k + 2) * w
6. :O only problem is, what the on earth is k. Perhaps you mean the constant? Or i think you mean x, the size of the stair etc.

However, when i use x, i have 1/6 * x * x+1 * x+2....

1/6 * 3 * 4* 5 = 10.

the number for 3 is infact 4, and 10 is for 4. If i use 4 as x, i get 20... the number for 5. Am i doing it wrong?
Thanks for the help so far johhny!
7. (Original post by sminggggg)
So infact, the formula for ANY stair, on ANY size grid is infact.... cue drum roll...

T = (x(x-1) over 2) * n + (1/6)k(k + 1)(k + 2) + (1/6)k(k + 1)(k + 2) * w
"x(x - 1) over 2" should be "x(x + 1) over 2".

x and k should be the same letter - they both represent the number of steps in the staircase. (So change the x's to k's or vice versa.)

The number _ in the formula is the sum of the first (k - 1) triangular numbers, not the sum of the first k triangular numbers. So "(1/6)k(k + 1)(k + 2)" should be "(1/6)(k - 1)k(k + 1)".

--

Sum of triangular numbers
1st triangular number = 1 = (1/6)*1*2*3
2nd triangular number = 3 = (1/6)*2*3*4 - (1/6)*1*2*3
3rd triangular number = 6 = (1/6)*3*4*5 - (1/6)*2*3*4
4th triangular number = 10 = (1/6)*4*5*6 - (1/6)*3*4*5
...
mth triangular number = (1/2)m(m + 1) = (1/6)m(m + 1)(m + 2) - (1/6)(m - 1)m(m + 1)

Sum of first m triangular numbers = (1/6)m(m + 1)(m + 2)

because everything else cancels (the two red terms cancel, the two green terms cancel, etc).
8. muah, i could kiss you. All so clear now, thanks man. 10/10
9. was asked today why the 1st triangle number = (1/6)*1*2*3

...why does it? :/
10. (Original post by sminggggg)
was asked today why the 1st triangle number = (1/6)*1*2*3

...why does it? :/
The general result is that the mth triangular number equals (1/6)m(m + 1)(m + 2) - (1/6)(m - 1)m(m + 1). [This answers your question because, replacing m by 1, it follows that the 1st triangular number is (1/6)*1*2*3 - (1/6)*0*1*2 = (1/6)*1*2*3.]

Proof of the general result:

(1/6)m(m + 1)(m + 2) - (1/6)(m - 1)m(m + 1)
= (1/6)m(m + 1)[(m + 2) - (m - 1)]
= (1/6)m(m + 1)*3
= (1/2)m(m + 1)
= mth triangular number

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