Solving Sum with C(m,k) - (1/2) * C(m, m/2)

[neginf]

Homework Statement

Sum over k from 0 to m of |k/m - 1/2| * C(m,k) where m is even to get (1/2) * C(m,m/2).

Homework Equations

C(m,k)=C(m,m-k)
1/2 - (m/2) / m = 0
(k/m) * C(m,k) = C(m-1,k-1)

The Attempt at a Solution

Tried things like:
sum over k from 0 to m of |k/m - 1/2| * C(m,k)
2 * sum over k from 0 to m/2 of |k/m - 1/2| * C(m,k)
= 2 * sum k from 0 to m/2 of (1/2 - k/m) * C(m,k)
= 2 * sum k from 0 to m/2 of ((1/2) * C(m,k) - k/m * C(m,k))
= sum k 0 to m/2 of C(m,k) - 2 * [sum k from 1 to m/2 of C(m-1,k-1)]

 

[neginf]

What am I doing wrong ?

sum over k from 0 to n of |k/n - 1/2|*C(n,k)
=sum over k from 0 to n of |k/n * C(n,k) - 1/2 * C(n,k)|
=|sum over k from 1 to n-1 of C(n-1,k-1) - sum over k from 0 to n of 1/2 * C(n,k)|
=|2^(n-1) - 1/2 * 2^n|
=0

The 1/2 * C(n,n/2) term I hope to show the whole sum is equal to is in the second sum that's in the | | signs.
1/2 * C(n,k)=1/2 * C(n,0) + .. .+ 1/2 *C(n,n/2) + ... + 1/2 * C(n,n).