On 29 Apr 00, at 2:41, [EMAIL PROTECTED] wrote:
> The sum of the following series, ((N)!) / ((k!)((N-k)!)) starting at k=1
> to k=(N), is equal to (2^N). If an odd counterexample is found,
Pascal's Triangle:
1
1 1
1 2 1
1 3 3 1
etc etc
where each element (except the 1 at the top) is the sum of the
numbers diagonally above it (blanks count as zero).
In this triangle, the kth element in row n is n! / (k! * (n-k)!).
Why? Because the number of ways of choosing k objects from n is
the same as the number of ways of choosing k objects from (n-1)
_plus_ the number of ways of choosing (k-1) objects from (n-1) -
which gives us the same construction rule as Pascal's Triangle.
It's obvious that the sum of the elements in each row is twice the
sum of the elements in the row above (from the construction
algorithm). Since the sum of the elements in the very first row = 1 =
2^0, it follows that the sum of the elements in _every_ row is a
power of 2 - specifically 2^(n-1), where n is the row number,
counting the very first row as row 1.
There are no counterexamples.
Unfortunately this does not constitute a proof of the non-existence
of odd perfect numbers!
Regards
Brian Beesley
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
