Posted by Denis Borris on September 18, 2002 at 21:43:00:
In Reply to: There is a numerical method - posted by Subhotosh Khan on September 18, 2002 at 15:56:52:
: 2
: ....10
: 12.......8
: ....18......0
: 30.......8
: ....26......0
: 56.......8
: ....34
: 90
Correct; but the table can be "horizontal" just as well:
2...12...30...56...90 ...
.10...18...26...34....
.....8.....8.....8....
(your standup presentation seems to only complicate things!)
The "starting" numbers are the pegs: 2, 10 and 8; works like this:
2n / 1! + 10(n)(n-1) / 2! + 8(n)(n-1)(n-2) / 3!
If there was a 4th line: ?(n)(n-1)(n-2)(n-3) / 4!
And this is what I meant by getting MESSIER: the longer the number of
steps to get to the common difference, the longer the equation...
The above simplifies to n(4n^2 + 3n - 1) / 3, which becomes the SUM
formula for the series; take n=3:
3(4*9 + 3*3 - 1) / 3 = 44; 2 + 12 + 30 = 44.