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n(n-1)
n:
1: 0
2
2: 2 2
4
3: 6 2
6
4: 12 2
8
5: 20 2
10
6: 30 2
12
7: 42 2
14
8: 56 2
16
9: 72
The numbers in the second column represent the difference between the numbers in the first column. And the numbers in the third column for the second. What is interesting is that 6+4+2=12, 12+6+2=20, 20+8+2=30 etc. If you add the numbers diagonally up you will get the next number.
I find that if you keep finding the differences over and over again, then add them up diagonally, going upwards, you will get the next number in the sequence
Another example
n^3-n^2-n
n:
1: -1
3
2: 2 10
13 6
3: 15 16
29 6
4: 44 22
51 6
5: 95 28
79
6: 174
As you get to the last column the difference decrease until you get the same number. Now, 44+29+16+6=95 and 95+51+22+6=174. Interesting huh?
It kinda reminds me of finding the derivative of the derivative of a distance time graph to get acceleration.
Can anyone tell me why this works?
Thanks.
Last edited by Toast (2007-02-19 20:37:52)
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A study of the first pattern reveals
n(n-1)-(n-1)(n-2)=n²-n-[n²-3n+2]=2n-2.
n 2n-2
1 0
2 2
3 6
4 8
5 10
The common difference in the terms is 2, initially, it appears to be an interesting pattern, mathematically though, quite obvious!
It would be unfair to deprive Toast of the analysis and the post.
Interesting observation, Toast!
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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I'll also post another conclusion I came to using the sequence in my Multiplication Squares problem:
(Discovered later)n:
1: 0
1
2: 1 2
3 1
3: 4 3
6 1
4: 10 4
10 1
5: 20 5
15
6: 35
As before, 10+6+3+1 = 20, 20+10+4+1 = 35 etc...
Let D_n = the number you want, evaluated by the formula
Hence, if you want to wanted to find what
would be, you would goOffline