Binomial Identity Proof Question (1 Viewer)

[adzy]

Can anybody do this question?

Show

(Nc1)^2 + 2(Nc2)^2 + 3(Nc3)^2 + .... n(NcN)^2 = (2n-1)! / [(n-1!)]^2

where NcR = N choose R

 

[tiggerfamilytre]

I think this is valid

 

[adzy]

Thanks, quite elegant.

 

[David_O]

Another method.

(nc1)^2 + 2(nc2)^2... + n(ncn)^2
= n/2 [ (nc0)^2 + (nc1)^2 ...+(ncn)^2) ] (since nCk = nC(n-k))
= n/2 . (2n)!/(n!)^2 (an almost standard result)
= n(2n-1)c(n-1) after some basic algebra.

tutor01: you knew what I meant p

 

[justchillin]

Tigger, you are either really skilled with computers or that is from another source...

 

[tiggerfamilytre]

not from another source, nor am i very skilled. Just used microsoft equation editor, then saved the equation box as an image in png form, then attached it to the post.

 

[ando_88]

oi tigger you bloody smart ass, you must of aced your 4unit trial ey? or is their some guy who always beats you? i hate those dam guys!

 

[tiggerfamilytre]

hey, you shut up .... i just hate them so much

 

[ando_88]

tigger, add me on msn man,

ando_simon@hotmail.com you won't be disapointed!

 

[justchillin]

I renounce my comments and bow to your computer mastery...