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ThudnBlunder
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 Combinatorial Sum   « on: Jan 20th, 2009, 5:09am » Quote Modify

49
Evaluate (-1)k 992k = 990 - 992 + 994 - ....... - 9998
k=0

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towr
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 Re: Combinatorial Sum   « Reply #1 on: Jan 20th, 2009, 7:50am » Quote Modify

1/2 sumn=0..99 C(99, n) in 1(99-n) + 1/2 sumn=0..99 C(99, n) (-i)n 1(99-n)
((1+i)99+(1-i)99)/2
[sqrt(2)99 exp(99 * 2pi * 1/8 i) + sqrt(2)99 exp(99 * 2pi * 7/8 i)]/2
248.5 [exp(6*pi/8 i) + exp(10*pi/8 i)]
248.5 * -sqrt(2)
-249
 « Last Edit: Jan 20th, 2009, 7:52am by towr » IP Logged

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pex
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 Re: Combinatorial Sum   « Reply #2 on: Jan 20th, 2009, 8:45am » Quote Modify

So, more generally, sumk=0..floor(n/2) (-1)k  nC2k =
 hidden: 2n/2 ......... if n =     0 mod 8      2(n-1)/2 ... if n = +- 1 mod 8      0 ............. if n = +- 2 mod 8    - 2(n-1)/2 ... if n = +- 3 mod 8    - 2n/2 ........ if n =      4 mod 8.

Interesting!
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