Posted by Subhotosh Khan on September 19, 2002 at 09:25:04:
In Reply to: Re: There is a numerical method - posted by Denis Borris on September 18, 2002 at 21:43:00:
: : 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.