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   Author  Topic: Combinatorial Sum  (Read 1581 times)
ThudnBlunder
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Combinatorial Sum  
« on: Jan 20th, 2009, 5:09am »
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             49
Evaluate (-1)k 992k = 990 - 992 + 994 - ....... - 9998  
            k=0
 
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Re: Combinatorial Sum  
« Reply #1 on: Jan 20th, 2009, 7:50am »
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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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Re: Combinatorial Sum  
« Reply #2 on: Jan 20th, 2009, 8:45am »
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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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