Maths Coursework Number Stairs. URGENT HELP NEEDED!

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16 years ago
#21
(Original post by 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.
Yeah, those workings out were horrible
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#22
(Original post by 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.
I have done 95% of the coursework on my own i just need the last formula!
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16 years ago
#23
(Original post by courtyboy02)
I have done 95% of the coursework on my own i just need the last formula!
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.
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16 years ago
#24
(Original post by 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*?
But what's teh difference between asking someone else in the class who's done it or another maths teacher?
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16 years ago
#25
(Original post by 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*?
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.
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16 years ago
#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.

(Original post by 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.
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.

(Original post by mik1a)
People are effectively changing the criteria of a maths GCSE from maths to knowledge of internet sites...
For example using maths reference sites: Dr maths, NRich. Mathworld, ...
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16 years ago
#27
(Original post by 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.
Hi mik1a, do you mean that we shouldn't really help with the course-work queries ???
Just give general hints and that's it ???
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16 years ago
#28
(Original post by 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.
We don't in fact need the formula for 1 + 4 + 9 + 16 + ... + n^2.

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.

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15 years ago
#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)
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15 years ago
#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
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15 years ago
#31
Bump
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15 years ago
#32
The six numbers are

n + 20
n + 10 ___ n + 11
n + 10 ___ n + 1 ____ n + 2

(n + 20) + (n + 10) + (n + 11) + n + (n + 1) + (n + 2)
= 6n + 20 + 10 + 11 + 1 + 2
= 6n + 44
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15 years ago
#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
=========
Oh man, I really need help on that, I remember my friend showing me that but I dont get it
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15 years ago
#34
Bump....please, i really need some help
0
15 years ago
#35
0
15 years ago
#36
B
U
M
P
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15 years ago
#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.
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15 years ago
#38
(Original post by Verycold)
Oh man, I really need help on that, I remember my friend showing me that but I dont get it
I've updated my original work. See if it helps any.
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