deskochan
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I have done the part (a) A(n) for even number but the odd number and all below I have no idea.
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mqb2766
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(Original post by deskochan)
I have done the part (a) A(n) for even number but the odd number and all below I have no idea.
Can you not just sketch a few cases like
2*2, 3*3, 4*4, 5*5, ...
to get the intuition? They've done that for you with an example when n is even.
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deskochan
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(Original post by mqb2766)
Can you not just sketch a few cases like
2*2, 3*3, 4*4, 5*5, ...
to get the intuition? They've done that for you with an example when n is even.
I just count the number of points and get the pattern of even the number of n of the triangle in the figure. 1,3,5, ... (n-5), (n-3). but the odd number I don't get it. Could you tell me more you mention 2*2... intuition? Because I just self-study pure mathematics.
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deskochan
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I get the linear programming constraint from the figure and part b. x>y, x+y>n, x<n and I think n = odd number, A(n) is in even number pattern, 2,4,6,...(n-?).
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mqb2766
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(Original post by deskochan)
I just count the number of points and get the pattern of even the number of n of the triangle in the figure. 1,3,5, ... (n-5), (n-3). but the odd number I don't get it. Could you tell me more you mention 2*2... intuition? Because I just self-study pure mathematics.
When n is odd, you should just sum the even numbers so 2+4+6+ ... The so-called rectangular numbers. It should be self evident from a sketch - upload what you've tried?

When n was even, you summed the odd numbers which is a square number?

I just meant drawing the different grids. So
n=1 is 2*2 which has 0 in the interior
n=2 is 3*3 which has 2 in the interior - the first rectangular number
n=3 is 4*4 which has 1+3 = 4 = 2^2 - summing odd numbers, or successive triangular numbers
n=4 is 5*5 which has 2+4 ....
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deskochan
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(Original post by mqb2766)
When n is odd, you should just sum the even numbers so 2+4+6+ ... The so-called rectangular numbers. It should be self evident from a sketch - upload what you've tried?

When n was even, you summed the odd numbers which is a square number?

I just meant drawing the different grids. So
n=1 is 2*2 which has 0 in the interior
n=2 is 3*3 which has 2 in the interior - the first rectangular number
n=3 is 4*4 which has 1+3 = 4 = 2^2 - summing odd numbers, or successive triangular numbers
n=4 is 5*5 which has 2+4 ....
At least I know, For n=even, A(n) is an odd number pattern, 1,3,5,7,... (n-5), (n-3); for n=odd, A(n) is an even number pattern 2,4,6,8,.. (n-5), (n-3). This is part (a) I get.
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mqb2766
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(Original post by deskochan)
At least I know, For n=even, A(n) is an odd number pattern, 1,3,5,7,... (n-5), (n-3); for n=odd, A(n) is an even number pattern 2,4,6,8,.. (n-5), (n-3). This is part (a) I get.
I take it you know the sum of those series? They're linear, so the sums are quadratics. n odd is a square number (sum of two successive triangular numbers) and n even is a rectangular number?
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deskochan
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(Original post by mqb2766)
When n is odd, you should just sum the even numbers so 2+4+6+ ... The so-called rectangular numbers. It should be self evident from a sketch - upload what you've tried?

When n was even, you summed the odd numbers which is a square number?

I just meant drawing the different grids. So
n=1 is 2*2 which has 0 in the interior
n=2 is 3*3 which has 2 in the interior - the first rectangular number
n=3 is 4*4 which has 1+3 = 4 = 2^2 - summing odd numbers, or successive triangular numbers
n=4 is 5*5 which has 2+4 ....
(Original post by mqb2766)
I take it you know the sum of those series? They're linear, so the sum is quadratics. n odd is a square number and n even is a rectangular number?
You are a genius and I just follow the pattern. For odd, A4, A6, A8 ,..= 1^2, 2^2, 3^2,....; For even, A5, A7, A9 = 2, 6, 12....
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Äries
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Hong Kong Advanced Level Pure Mathematics past examination paper.
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deskochan
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(Original post by Äries)
Hong Kong Advanced Level Pure Mathematics past examination paper.
Yes, 1984
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mqb2766
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(Original post by deskochan)
You are a genius and I just follow the pattern. For odd, A4, A6, A8 ,..= 1^2, 2^2, 3^2,....; For even, A5, A7, A9 = 2, 6, 12....
Tbh, the basic triangular numbers, rectangular numbers, square numbers patterns are fairly well known and easy to derive. This part of the question is just a slightly different way to ask the same thing.
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deskochan
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(Original post by mqb2766)
Tbh, the basic triangular numbers, rectangular numbers, square numbers patterns are fairly well known and easy to derive. This part of the question is just a slightly different way to ask the same thing.
Examination always does the same thing in various ways. I am a layman to follow the road of this world.
By the way, A4+A6+A8.. = from summation of (n-1)^2 (from n=2 to ?) , A5+A7+A9.. = summation of n(n-1) (from n=2 to?).
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mqb2766
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(Original post by deskochan)
Examination always does the same thing in various ways. I am a layman to follow the road of this world.
By the way, A4+A6+A8.. = from summation of (n-1)^2 (from n=2 to ?) , A5+A7+A9.. = summation of n(n-1) (from n=2 to?).
Not sure what you mean.
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deskochan
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(Original post by mqb2766)
Not sure what you mean.
I mean For n=odd A4+ A6,+A8 +....= 1^2,+2^2,+3^2,....; For n=even, A5 + A7 + A9 = 2 + 6 +12...
I go to part b(ii) find B(2k). I think B(2k) is added the above two series together. Is this logic correct?
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mqb2766
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(Original post by deskochan)
I mean For n=odd A4+ A6,+A8 +....= 1^2,+2^2,+3^2,....; For n=even, A5 + A7 + A9 = 2 + 6 +12...
I go to part b(ii) find B(2k). I think B(2k) is added the above two series together. Is this logic correct?
I agree that An is a sequence of square numbers and rectangular numbers.
Which part of the question are you doing now.
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deskochan
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(Original post by mqb2766)
I agree that An is a sequence of square numbers and rectangular numbers.
Which part of the question are you doing now.
Part b. Thank you for your help and I have finished part (a)
In part (a), I get n=even A(n) = [(n-2)/2]^2, n=odd, A(n) = (n-1)/2 * (n-3)/2
For part b(i), I think the non-degenerate triangle is x+y>t (from the diagram, x=n is the longest side of the triangle), and it is given y<x<t = n and there are the three inequalities, x>y, x+y>t and x<t and thus the number of triangles is equal to the number of integral points A(n) in part (a).
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mqb2766
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(Original post by deskochan)
Part b. Thank you for your help and I have finished part (a)
In part (a), I get n=even A(n) = [(n-2)/2]^2, n=odd, A(n) = (n-1)/2 * (n-3)/2
For part b(i), I think the non-degenerate triangle is x+y>t (from the diagram, x=n is the longest side of the triangle), and it is given y<x<t = n and there are the three inequalities, x>y, x+y>t and x<t and thus the number of triangles is equal to the number of integral points A(n) in part (a).
Yes, agree with that.
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deskochan
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I go to part b (ii), I think B(2k) = A4+A5+....... and I need to add two series: even (square pattern number) and odd (rectangle pattern number).
Thus, I try to add B(2k) = summation A(i) (from i=4 to 2k).
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mqb2766
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(Original post by deskochan)
I go to part b (ii), I think B(2k) = A4+A5+....... and I need to add two series: even (square pattern number) and odd (rectangle pattern number).
Thus, I try to add B(2k) = summation A(i) (from i=4 to 2k).
Just make sure you're doing the first few numbers correctly. The sum of square number and rectangular numbers are not too hard. Obviously, they must form a cubic, as the diesired expression suggests.
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deskochan
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(Original post by mqb2766)
Just make sure you're doing the first few numbers correctly. The sum of square number and rectangular numbers are not too hard. Obviously, they must form a cubic, as the diesired expression suggests.
I get stuck in this part and guide me more, please.
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