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    ok, im retaking this subject and already started my 1st piece of coursework, as I went through this, i do not understabd :confused:

    Please can someone go through this with me

    I've worked out a formula for any size grid with any size stair/step-size.

    Let the grid size be G
    Let the step size be S
    Let the number in the starting cell be N

    Along the first row, the sum of the S numbers is

    S1 = N + (N+1) + (N+2) + ... + (N+S-1)


    S1 = ∑n {n=N to N+S-1}

    Similarly, in the kth row,

    Sk = ∑n {n=N+G(k-1) to N+G(k-1)+S-k}

    The starting number in each succesive row increases by G and the number of terms in each row decreases by 1.

    There are S rows, So the total sum is,

    St = ∑Si {i=1 to S}


    Sk = ∑n {n=N+G(k-1) to N+G(k-1)+S-k}


    Sk = ∑n {n=1 to N+G(k-1)+S-k} - ∑n {n=1 to N+G(k-1)-1}

    Using the standard formula for the sum of r terms, viz. ∑ i {r=1 to r} = ½r(r+1),

    Sk = ½(N+G(k-1)+S-k)(N+G(k-1)+S-k+1) - ½(N+G(k-1)-1)(N+G(k-1))

    I'll let you work this out (watch your arithmetic!), but it comes to,

    Sk = ½{(S+1)(2N-2G+S) - k(2N-4G-2GS+2S+1)-k²(2G-1)}
    ================================ ==========


    St = ∑Si {i=1 to S}

    St = ½S(S+1)(2N-2G+S) - ½(2N-4G-2GS+2S+1)∑ i {i=1 to S} - ½(2G-1)∑ i² {i=1 to S}

    St = ½S(S+1)(2N-2G+S) - ¼(2N-4G-2GS+2S+1)S(S+1) - (1/12)(2G-1)S(S+1)(2S+1)
    ================================ ============================

    Putting G= 10, and S = 3,

    St = ½.3(4)(2N - 20 + 3) - ¼(2N - 40 - 60 + 6 + 1).3.(4) - (1/12)(20-1).3.(4)(7)
    St = 12N - 102 - 6N + 279 - 133
    St = 6N + 44

    Just to check how far you've got, do you know how to get these formulas?

    2-step stairs on 10-by-10 grid:
    3N + 11

    3-step stairs on 10-by-10 grid:
    6N + 44

    4-step stairs on 10-by-10 grid:
    10N + 110

    S-step stairs on a 10-by-10 grid:
    [(N - 1)S(S + 1)/2] - (4/3)S + (1/2)S^2 + (11/6)S^3

    The material you quoted is advanced. You probably don't need to include it in your project even to get an A*.
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