# Maths c/w - number stairs

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#1
Hello i really need sum help getting the general formula of any grid size, any step stair, of any starting point. i have found a few formulas which i need to use quadratic and cubic equations. Can anyone show me the working out to get to the final formula plz
thanks
kush
0
17 years ago
#2
(Original post by Kush)
Hello i really need sum help getting the general formula of any grid size, any step stair, of any starting point. i have found a few formulas which i need to use quadratic and cubic equations. Can anyone show me the working out to get to the final formula plz
thanks
kush
I'm not familiar with this piece. Could you just give me a brief overview of what you've done?
0
#3
(Original post by meepmeep)
I'm not familiar with this piece. Could you just give me a brief overview of what you've done?
0
#4
(Original post by meepmeep)
I'm not familiar with this piece. Could you just give me a brief overview of what you've done?
i have got a few formulas:
(1step) x
(2step) 3x + G + 1
(3step) 6x + 4G + 4
(4step) 10x + 10G + 10
(5step) 15x + 20G + 20
(6step) 21x + 35G + 35
(7step) 28x + 56G + 56

where x is starting point, G is grid size

i need to use quadratics and cubics here. i found the differences between these and the x values are constant at 2nd difference which means quadratics need to be used. For the G terms the constant difference comes at the 3rd difference so cubic needs to be used. The constant 4 G term is G+1 and according to cubics G+1 = 6a
a=G+1 / 6
but i cant work out b,c,d. plz help
0
17 years ago
#5
(Original post by Kush)
i have got a few formulas:
(1step) x
(2step) 3x + G + 1
(3step) 6x + 4G + 4
(4step) 10x + 10G + 10
(5step) 15x + 20G + 20
(6step) 21x + 35G + 35
(7step) 28x + 56G + 56

where x is starting point, G is grid size

i need to use quadratics and cubics here. i found the differences between these and the x values are constant at 2nd difference which means quadratics need to be used. For the G terms the constant difference comes at the 3rd difference so cubic needs to be used. The constant 4 G term is G+1 and according to cubics G+1 = 6a
a=G+1 / 6
but i cant work out b,c,d. plz help
Alright, if you look at the progression that occurs in the x:
1, 3, 6, 10, 15..., these are triangle numbers. The formula for the nth triangle number is n(n+1)/2, so with n steps, the first part will be:
n(n+)x/2

I'll show you two ways to tackle it.

The difference method, as you did before:
n.......1..........2.........3...........4.............5...............6
term...0..........1.........4..........10...........20.............35
1stdif.........1........3.........6............10...........15
2nddif............2........3...........4.............5
3rddiff................1..........1............1

Therefore, as you stated, if n is the number of steps, (1/6)n^3 is the first term.

Put this in:
n.............1..........2.........3...........4.............5...............6
term.........0..........1.........4..........10...........20.............35
(1/6)n^3...(1/6)....(8/6)....(27/6)...(64/6)......(125/6)........(21/6)
diff...........1/6......2/6.....3/6......4/6.........5/6...........1
1stdiff...........1/6......1/6.......1/6........1/6..........1/6

This gives -n/6
(1/6)n^3-n/6, which works. So stick this before the G+1.

Other way to tackle it (and you can prove it this way), is to use sigma notation:

If the term is found by adding the triangle numbers up to the (n-1)th triangle number.

This is equal to:

1x2/2+2x3/2+3x4/2+......+(n-1)n/2

This sum can be represented by the following:

Sigma(between r=1 and r=n-1) r(r-1)/2

Sigma is represented by the sigma sign which looks a bit like an E, so I'll use that to represent it:

E r(r-1)/2 = (1/2)E r(r-1)
=(1/2)E r^2-r
=(1/2)(E r^2 - E r)

Now, sigma between r=1 and r=n, the total for r is n(n+1)/2 and for r^2 is n(n+1)(2n+1)/6
However, you need to replace n with n-1, giving the sum of r to be n(n-1)/2 and the sum of r^2 to be (n-1)n(2n-1)/6

Times each of these by a half and you get n(n-1)/4 and (n-1)n(2n-1)/12

Add these together and you get (3n(n-1)+(n-1)n(2n-1))/12=n(n-1)(2n-1+3)/12=n(n-1)(2n-1)/12=(2n^3-2n)/12=n^3/6+n/6
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