# Maths Coursework Number Stairs. URGENT HELP NEEDED!

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#1
What is the formula to work out the total of any sized step stair on any grid in any position. I also need the proof that the formula works. PLEASE HELP I AM REALLY STUCK. I will help any1 who needs help with science coursework if they help me!
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15 years ago
#2
can you post the full question, then someone may be able to help you.
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#3
Can u simplify this formula.

(m/2 + 3m/2 + 1) + (1/6m3 + 1/2m2 + 1/3m) + (1/6m3 + 1/2m2 + 1/3m)
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15 years ago
#4
(Original post by courtyboy02)
Can u simplify this formula.

(m/2 + 3m/2 + 1) + (1/6m3 + 1/2m2 + 1/3m) + (1/6m3 + 1/2m2 + 1/3m)
does 1/6m3 mean one sixth lots of m cubed? i don't really get your notation as such.
if so, then you will get

2m + 1 + 1/3 m^3 + m^2 + 2/3 m.
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#5
(Original post by 4Ed)
does 1/6m3 mean one sixth lots of m cubed? i don't really get your notation as such.
if so, then you will get

2m + 1 + 1/3 m^3 + m^2 + 2/3 m.
Yes it is cubed. couldn't use subscript on ehre. Is that correct then?
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15 years ago
#6
What is m?

We need to know (1) the size of the stairs and (2) their position. So shouldn't there be two variables?
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15 years ago
#7
OK, if you're doing the Number Stair Coursework for Maths, then the formula I found will work for any sized number grid:

6N + 44

N= (Bottom Left) Number you start the grid at.

For this project I got an A*, and it works.
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15 years ago
#8
Go on:

www.coursework.info

and research some ideas .
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15 years ago
#9
(Original post by Stumbleines)
the formula I found will work for any sized number grid:

6N + 44

N= (Bottom Left) Number you start the grid at.
That works for three step number stairs (see http://www.uk-learning.net/t51696.html) but courtyboy02 wants a formula for number stairs of all sizes.
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#10
Jonny W is right. I already know the formulas for the 3 step number stair and all the number stairs. I am looking for the formula which will calculate the total of any sized step stair on any sized grid in any position.
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#11
Jonny W the link u gave me only tells me the formula for any 3 step stair in any position.
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15 years ago
#12
Let N be the position of the stairs (the number under the bottom-left square).

Let S be the number of steps in the stairs.

Let f(N, S) be the total under the stairs for position N and size S. Our task is to work out f(N, S) for every value of N and S.

f(1, 4) is the sum of this:

31
21 22
11 12 13
01 02 03 04

f(64, 4) is the sum of this:

94
84 85
74 75 76
64 65 66 67

Each number has gone up by 63. Therefore the sum has gone up by 63*(number of squares in the stairs) = 63 * 10 = 630. So we have proved that f(64, 4) = 630 + f(1, 4).

For the same reason,

f(N, S)
= (N - 1)*(number of squares in S-step stairs) + f(1, S)
= (N - 1)*(1 + 2 + 3 + ... + S) + f(1, S)
= [(N - 1)S(S + 1)/2] + f(1, S).

Our task is now easier because we only have to work out f(1, S) for each value of S. We can do that by the usual method of calculating differences.

f(1, 1) = 1
f(1, 2) = 14
f(1, 3) = 50
f(1, 4) = 120
f(1, 5) = 235
f(1, 6) = 406
f(1, 7) = 644
f(1, 8) = 960

1st differences:
13, 36, 70, 115, 171, 238, 316

2nd differences:
23, 34, 45, 56, 67, 78

3rd differences:
11, 11, 11, 11, 11

I'll leave this bit to you, but I make the answer to be

f(1, S) = - (4/3)S + (1/2)S^2 + (11/6)S^3.

So, using the previous red formula,

f(N, S) = [(N - 1)S(S + 1)/2] - (4/3)S + (1/2)S^2 + (11/6)S^3.

As a check, let's put S = 3 and see if we get the formula for 3-step stairs.

f(N, 3)
= [(N - 1)*3*4/2] - (4/3)*3 + (1/2)*9 + (11/6)*27
= 6(N - 1) - 4 + (9/2) + (99/2)
= 6N - 6 - 4 + (9/2) + (99/2)
= 6N + 44.
0
15 years ago
#13
Jonny W has given you some excellent help with this coursework. What he's has done so far covers the 3-step stair and also any size stair, but in a 10x10 grid only.

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
=========
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#14
Thank you very much for ur help. What do u mean by the kth row, is it similar to the term, 'n'?
Thanks again
Courtyboy
0
15 years ago
#15
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.
0
15 years ago
#16
(Original post by courtyboy02)
Thank you very much for ur help. What do u mean by the kth row, is it similar to the term, 'n'?
Thanks again
Courtyboy
The kth row is just a general row, somewhere in the stair, used to get a general result/formula.

Suppose you have a 5-step stair, a grid size of 10 and starting at N=25.

The first row is 25 + 26 + 27 + 28 + 29
The 2nd row is 35 + 36 + 37 + 38
.
.
.
The kth row is (25+10(k-1)) + (25+ 10(k-1) + 1) + ...

n is just used as a counting number, the same as i,k or r.
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15 years ago
#17
(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.
There can be a fine line between helping someone and doing the work for them. But then anyone can get the same help here - they just have to ask.
Anyway, courtboy still has to do some work. He's still got this to do,

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)}
================================ ==========
0
15 years ago
#18
(Original post by Fermat)
There can be a fine line between helping someone and doing the work for them. But then anyone can get the same help here - they just have to ask.
Anyway, courtboy still has to do some work. He's still got this to do,
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. People are effectively changing the criteria of a maths GCSE from maths to knowledge of internet sites...

crazy cheaters
0
15 years ago
#19
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.
0
15 years ago
#20
(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. People are effectively changing the criteria of a maths GCSE from maths to knowledge of internet sites...

crazy cheaters
Amen..........i hate f*****g cheaters......this thread reminds me of P2 and Pete2004???????? No wonder this sites going down hill.........it lets people cheat and the mods do nothing!
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