Email address obfuscation in effect -- please
click here to turn it off.
[
Date Prev][
Date Next][
Thread Prev][
Thread Next][
Date Index][
Thread Index]
On Fri, 18 Apr 2008, Mike Miller wrote:
This is what I get:
N
Sum (N choose i) [2^(i-1)]
i=1
(But I now realize that adds up to half of one less than 3^N ; see below.)
It turns out that this identity is almost obvious when I put it in the
right form. Here's the form:
N
Sum (N choose i) 2^i = 3^N
i=0
Then I see that it's just the expansion of (2 + 1)^N. Ha!
I like to do a little math review now and then so that I don't forget
everything I once knew.
Mike
_______________________________________________
members mailing list
EMAIL:PROTECTED
http://mlug.missouri.edu/mailman/listinfo/members
Home | FAQ | Server | Presentations | Mailing Lists/Archives | Member Tools | Links | Sponsors | Contact