# Maths Coursework Number Stairs. URGENT HELP NEEDED!

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#21

(Original post by

We can use

f(N, S, G) = [(N - 1)S(S + 1)/2] + f(1, S, G)

to simplify the calculations in Fermat's method.

**Jonny W**)We can use

f(N, S, G) = [(N - 1)S(S + 1)/2] + f(1, S, G)

to simplify the calculations in Fermat's method.

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(Original post by

I cannot believe what i see before me..............coursework you are supposed to do on your own!!!!!!!!!!!! This is blantent cheating i did my coursework on my own so why the hell cant others..............this is an absolute disgrace.

**Economics**)I cannot believe what i see before me..............coursework you are supposed to do on your own!!!!!!!!!!!! This is blantent cheating i did my coursework on my own so why the hell cant others..............this is an absolute disgrace.

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#23

(Original post by

I have done 95% of the coursework on my own i just need the last formula!

**courtyboy02**)I have done 95% of the coursework on my own i just need the last formula!

Sorry it's frank but it's how the system should work.

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#24

(Original post by

Yes the last formula is pribably the distinction between an A and an A*... if you can't do it then why shopuld you get the A*?

**mik1a**)Yes the last formula is pribably the distinction between an A and an A*... if you can't do it then why shopuld you get the A*?

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#25

**mik1a**)

Yes the last formula is pribably the distinction between an A and an A*... if you can't do it then why shopuld you get the A*?

1 + 4 + 9 + 16 + ... + n^2,

which isn't taught at GCSE level, it seems to me that almost everyone who completes the investigation will have had help from someone.

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#26

Cheating is frowned upon of course, but how can you tell when someone just can't be bothered to do their own work, and others who simply need a lot of help???

I try to give a mixture of solving problems and giving hints.

If I see someone always asking for help, who just puts forward a question and says "how do you do this?" and that's it, then I end up ignoring them.

There are various ways that people can do course-work. They can do it all by themselves using their own intrinsic knowledge, they can research, on their own, and get help that way; they can ask for help, from teachers, friends (and, if they're lucky) their parents. They can use their own talent at maths to complete the course-work or they can use what talent they have at getting information from other sources, whatever they may be.

For example using maths reference sites: Dr maths, NRich. Mathworld, ...

I try to give a mixture of solving problems and giving hints.

If I see someone always asking for help, who just puts forward a question and says "how do you do this?" and that's it, then I end up ignoring them.

(Original post by

Not everyone can ask though, not everyone that is doing the coursework. If you give him help then that is help that other people aren't getting.

**mik1a**)Not everyone can ask though, not everyone that is doing the coursework. If you give him help then that is help that other people aren't getting.

(Original post by

People are effectively changing the criteria of a maths GCSE from maths to knowledge of internet sites...

**mik1a**)People are effectively changing the criteria of a maths GCSE from maths to knowledge of internet sites...

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#27

(Original post by

Yes the last formula is pribably the distinction between an A and an A*... if you can't do it then why shopuld you get the A*?

Sorry it's frank but it's how the system should work.

**mik1a**)Yes the last formula is pribably the distinction between an A and an A*... if you can't do it then why shopuld you get the A*?

Sorry it's frank but it's how the system should work.

Just give general hints and that's it ???

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#28

(Original post by

Since getting the formula for f(N, S, G) requires knowledge of the formula for

1 + 4 + 9 + 16 + ... + n^2,

which isn't taught at GCSE level, it seems to me that almost everyone who completes the investigation will have had help from someone.

**Jonny W**)Since getting the formula for f(N, S, G) requires knowledge of the formula for

1 + 4 + 9 + 16 + ... + n^2,

which isn't taught at GCSE level, it seems to me that almost everyone who completes the investigation will have had help from someone.

Let f(N, S, G) = total under the stairs for position N and size S on a G by G grid.

Using the method in http://www.uk-learning.net/showthrea...620#post955620 we can obtain formulas for f(N, S, 1), f(N, S, 2), f(N, S, 3), and so on.

f(N, S, 1) = [(N - 1)S(S + 1)/2] + (1/6)S + (1/2)S^2 + (1/3)S^3,

f(N, S, 2) = [(N - 1)S(S + 1)/2] + (0)S + (1/2)S^2 + (1/2)S^3,

f(N, S, 3) = [(N - 1)S(S + 1)/2] + (-1/6)S + (1/2)S^2 + (2/3)S^3,

f(N, S, 4) = [(N - 1)S(S + 1)/2] + (-1/3)S + (1/2)S^2 + (5/6)S^3,

f(N, S, 5) = [(N - 1)S(S + 1)/2] + (-1/2)S + (1/2)S^2 + (1)S^3,

f(N, S, 6) = [(N - 1)S(S + 1)/2] + (-2/3)S + (1/2)S^2 + (7/6)S^3,

f(N, S, 7) = [(N - 1)S(S + 1)/2] + (-5/6)S + (1/2)S^2 + (4/3)S^3.

The green numbers are always 1/2. The red numbers go down by 1/6 between consecutive lines. The blue numbers go up by 1/6 between consecutive lines. So

red number on line G = 1/3 - G/6,

green number on line G = 1/2,

blue number on line G = 1/6 + G/6.

So the general formula is

f(N, S, G) = [(N - 1)S(S + 1)/2] + (1/3 - G/6)S + (1/2)S^2 + (1/6 + G/6)S^3.

This answer agrees with Fermat's.

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#29

i was given this formula by an anonamous guest

'Ah well, i had a sudden idea so i decided to do it myself without help. If anyone is interested, i ended up with

T=(P²+P)/2)S + (P³-P)/6)W) + (P³-P)/6)

T being the sum of each square inside the stairs

S being the s number, the lower left square.

P being the width of a stair (must be height and width)

W being the height and width of the grid'

it works, as you can put it in any size grid, any size step and any place on a grid!

but the only thing is i dont know how to get it, can anyone help, reply me at [email protected]

(also is anyone after a gmail account, 1gb free space and loads of new feature)

'Ah well, i had a sudden idea so i decided to do it myself without help. If anyone is interested, i ended up with

T=(P²+P)/2)S + (P³-P)/6)W) + (P³-P)/6)

T being the sum of each square inside the stairs

S being the s number, the lower left square.

P being the width of a stair (must be height and width)

W being the height and width of the grid'

it works, as you can put it in any size grid, any size step and any place on a grid!

but the only thing is i dont know how to get it, can anyone help, reply me at [email protected]

(also is anyone after a gmail account, 1gb free space and loads of new feature)

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#30

44

34 35

24 25 26

Total = 188

I know the formula is 6n+44 but im showing that I picking random 3 step, its just i need help for that formula ..i tried 6n+176 but doesnt equal 188, any suggestions, thanks

34 35

24 25 26

Total = 188

I know the formula is 6n+44 but im showing that I picking random 3 step, its just i need help for that formula ..i tried 6n+176 but doesnt equal 188, any suggestions, thanks

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#32

The six numbers are

n + 20

n + 10 ___ n + 11

n + 10 ___ n + 1 ____ n + 2

Adding these up gives

(n + 20) + (n + 10) + (n + 11) + n + (n + 1) + (n + 2)

= 6n + 20 + 10 + 11 + 1 + 2

= 6n + 44

n + 20

n + 10 ___ n + 11

n + 10 ___ n + 1 ____ n + 2

Adding these up gives

(n + 20) + (n + 10) + (n + 11) + n + (n + 1) + (n + 2)

= 6n + 20 + 10 + 11 + 1 + 2

= 6n + 44

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#33

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

=========

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

=========

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#37

I know the formula is:

(½s x (s+1))n + ((g+1)/6) x (s³-s))

s = step stair size

n = bottom left hand corner of stair

g = grid size

but im not sure how i got to it

i was given a formula but it didn't work. i worked out what was missing and then i found out that i could simplyfy it.

can anyone help me, if not, don't bother trying to say anything.

(½s x (s+1))n + ((g+1)/6) x (s³-s))

s = step stair size

n = bottom left hand corner of stair

g = grid size

but im not sure how i got to it

i was given a formula but it didn't work. i worked out what was missing and then i found out that i could simplyfy it.

can anyone help me, if not, don't bother trying to say anything.

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#38

(Original post by

Oh man, I really need help on that, I remember my friend showing me that but I dont get it

**Verycold**)Oh man, I really need help on that, I remember my friend showing me that but I dont get it

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