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Mike Miller wrote:
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.
Here is a math problem I worked on when I was a student. Show that
N
Sum (N choose i) x(x+i)^(i-1) y(y+n-i)^(n-i) = (x+y)(x+y+n)^(n-1).
i=0
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