Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 #1
    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. jcsd
  3. Jun 25, 2011 #2
    Theese two formulas should be useful:
    [itex]\sum_{k=0}^nC(n,k)=2^n[/itex]
    [itex]\sum_{k=0}^nkC(n,k)=n2^{n-1}[/itex]
    Does it help?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: 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. M - (x -n)^2 = ? (Replies: 16)

  2. Egyptian 2/n-table (Replies: 10)

  3. Mth digit of 2^n (Replies: 10)

Loading...