- #1
neginf
- 56
- 0
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 ?
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 ?