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    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 :confused:

    All help much loved and apreciated, Sean
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    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".

    Look at this thread: http://www.thestudentroom.co.uk/t54167.html
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    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
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    (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
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    !! 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
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    :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!
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    (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)

    Adding up,

    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).
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    muah, i could kiss you. All so clear now, thanks man. 10/10
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    was asked today why the 1st triangle number = (1/6)*1*2*3

    ...why does it? :/
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    (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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