Katie Wesley
Posted on Sunday, 25 May, 2003 - 12:55 pm:
Can anybody think of how to go about proving that:
2k
å
i=0 
 (-1)i (2kCi2) = k
å
i=0 
 (kCi)2
Katie Wesley
Posted on Sunday, 25 May, 2003 - 01:52 pm:
By the way the original question was from STEP II 1999 Question 4(i)and was:
Show that, if n is an even integer, then
n
å
i=0 
 (-1)i (nCi)2 = (-1)n/2 (nCn/2)


I rewrote both sides with n=2k and then (after expansion using the binomial theorem) considered the coefficients of xk on both sides of the identity
(1-x)2k = (1-x)k (1-x)k

The coefficient of xk in (1-x)2k is (-1)k (2k Ck ) which is the same as the right hand side of the rewritten version of the question. Therefore the coefficient of xk in (1-x)k (1-x)k (which is the right hand side in my previous message, after using the fact that k Cr =k Ck-r ) must be equal to the left hand side of the rewritten question.

This has left me trying to answer the question I posed in the previous message. My teacher couldn't think of how to do it and I can't so could someone please help? It's driving me mad!
Kerwin Hui
Posted on Sunday, 25 May, 2003 - 02:49 pm:
The question is: Prove that
Latex image click or follow link to see src
which is immediate from considering the constant term in the identity
Latex image click or follow link to see src

Kerwin
Katie Wesley
Posted on Sunday, 25 May, 2003 - 03:44 pm:
Thanks, how did you come up with that identity? It just seems to have been plucked from thin air!
I have trouble thinking of identities that will help me arrive at a solution - any suggestions?
Kerwin Hui
Posted on Sunday, 25 May, 2003 - 05:09 pm:
The (-1)j in the left hand side gives a clue - you want to consider something of the form (x+y)2k (z-w)2k . The right hand side suggests you need something of the form (x-y)k (z-w)k . Now there are too many variables, so why not set some of them to be 1 and see what happens? Setting y=z=1, the LHS wants something of the form
(1+x)2k (1-w)2k
and there are still too many variables! Set x=w, we get
(1+x)2k (1-x)2k
of which the coefficient of x2k is the one we want. Now this expression is also equal to (1-x2 )2k which can be written in the form the RHS wants. Finally, I just divided the identity by x2k (treating x as an indeterminate rather than a number) to introduce a little bit of symmetry, but that is not necessary.

Kerwin