20 Aug
2014
20 Aug
'14
8:55 p.m.
sum( (D+k-j)! / (k-j)! * (-1)^j * (2*k+1+D-2*j)^(2*k+1) / (j! * (2*k+1+D-j)!), j, 0, k ) = 4^k claims lots of checking. (only checked when D>0 is integer.) But... how to prove it? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)