- #1

neginf

- 56

- 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 ?