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# Number stairs watch

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

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)

or

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}
=============

Now,

Sk = ∑n {n=N+G(k-1) to N+G(k-1)+S-k}

or

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)}
================================ ==========

And,

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
=========
2. 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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