Alternating binomial sum

sevvost · April 28th, 2006, 7:09 am

Evaluate

pk14 · April 28th, 2006, 9:07 pm

When n is odd, clearly the sum equals to 0.When n is even, after some calculation (I may type up later), I got the value is Hope it is correct.

sevvost · April 28th, 2006, 9:17 pm

Yes, this is correct. I am curious in your approach, could you please share. No need to type all calculations, I know this dismal equation editor is a pain.

pk14 · April 28th, 2006, 9:48 pm

Thank you sevvost! Below is what I did. It looks confusing but actually very easy^_^.Let's denote the sum we want to evalue . Since we know when n is odd, the sum is zero, we may assume n is an even number from now on.Clearly, we have,and which gives us Time (n+2) to the first equation and substract the third equation, we get,Hence,

sevvost · April 28th, 2006, 10:27 pm

Cool. Another approach is to use the identity (= B(a+1, b+1) ). Therefore we have ( everywhere):

pk14 · April 30th, 2006, 4:05 am

sevvost,Very beautiful! Better than what I did!Also, thank you for questions!

sevvost · April 30th, 2006, 4:20 am

Thank you for your kind comments. Yes, it is a nice trick (did not invent it myself, of course.) I also thought it might be fun to shift gears once in a while

wizwx · May 22nd, 2006, 9:07 pm

Why the sum is clearly zero when n is odd?When n is odd, the ith item and the (n-i)th item cancel each other, but that leaves the item (n+1)/2, right?

vixen · May 22nd, 2006, 9:13 pm

QuoteOriginally posted by: wizwxWhy the sum is clearly zero when n is odd?When n is odd, the ith item and the (n-i)th item cancel each other, but that leaves the item (n+1)/2, right?The midpoint lies between items (n-1)/2 and (n+1)/2.

sevvost · May 22nd, 2006, 9:27 pm

I think the best way to settle questions like this is to do some experiments with small n.