# Sum over k from 0 to an even n of abs(k/n - 1/2)*C(n,k)=(1/2)*C(n,n/2)

1. Jun 24, 2011

### neginf

Odd number of terms so there is a middle term. It is equal to 0.
Since C(n,k)=C(n,n-k), abs(k/n - 1/2)=abs([n-k]/n - 1/2), and what's in the abs() increases and is positive k goes from 0 to n/2 -1,
the sum is twice the sum of the first n/2 - 1 terms, 2*(1/2 - k/n)*C(n,k).
Added C(n,k) from 0 to n, C(n-1,k-1) from 1 to n-1 seperately. Still do not see how to get.
Specific examples are equal to (1/2)*C(n,n/2) but how can this be shown for any even n ?

2. Jun 25, 2011

### Petr Mugver

Theese two formulas should be useful:
$\sum_{k=0}^nC(n,k)=2^n$
$\sum_{k=0}^nkC(n,k)=n2^{n-1}$
Does it help?