# Maths c/w - number stairs

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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

thanks

kush

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#2

(Original post by

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

**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

0

**meepmeep**)

I'm not familiar with this piece. Could you just give me a brief overview of what you've done?

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**meepmeep**)

I'm not familiar with this piece. Could you just give me a brief overview of what you've done?

(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

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#5

(Original post by

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

**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

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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