Prove Summation Equation: (2^n-1)n for Any Integer n

[tiny-tim]

Prove, for any integer n:

\sum_{0\,\leq\,m\,< n/2}\,(-1)^m(n - 2m)^n\,^nC_m\ =\ 2^{n-1}\,n!

for example, 7⁷ - 5⁷7 + 3⁷21 - 1⁷35

= 823543 - 546875 + 45927 - 35 = 304560 = 64 times 5040

 

[Gokul43201]

I'm guessing there is a brute force way to prove this as well as a clever combinatoric argument, and you're hoping we only find the first, so you can dazzle us with the second?

 

[tiny-tim]

I'm guessing there is a brute force way to prove this as well as a clever combinatoric argument, and you're hoping we only find the first, so you can dazzle us with the second?

Hi Gokul!

Sort of … I accidentally found a geometric-cum-combinatoric proof of this while looking at a homework thread,

but I couldn't help thinking that there must be some way of solving this just by looking at it and coming up with a solution …

but no ordinary technique comes to mind since the exponand (is that the right word? ) keeps changing.

I was hoping somebody knew a finding-the-solution technique (maybe for a simpler problem), rather than an already-knowing-what the-solution-is technique!

 

[Count Iblis]

Isn't this an exercise from this book:


http://www.math.upenn.edu/~wilf/AeqB.html