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Topic: Combinatorial sum (Read 419 times) |
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NickH
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Combinatorial sum
« on: Jun 7th, 2003, 2:21am » |
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Find a closed form expression for the sum, from k = 1 to n, of C(n,k) * k5, where C(n,k) = n! / [k! (n-k)!] is a binomial coefficient.
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TenaliRaman
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Re: Combinatorial sum
« Reply #1 on: Jun 8th, 2003, 12:11pm » |
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i will use this notation for summation,
Eni=0f(i)
The above summation given is,
Enk=1C(n,k)k5
I have converted this into a very difficult summation which is,
Enk=1(2n-[Ek-1i=0C(n,i)])(k5-[k- 1]5)
If the brackets are opened up,then it somewhat simplifies to,
2nn5-Enk=1[Ek-1i=0C(n,i)](k5-[k-1] 5)
This is as far as i could go.But i am pretty sure of two things,
1>its a equation of degree 5
2>the final answer contains 2n as factor
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ThudnBlunder
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Re: Combinatorial sum
« Reply #2 on: Jun 8th, 2003, 1:06pm » |
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3>the final answer contains that which I am trying to find.
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TenaliRaman
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Re: Combinatorial sum
« Reply #3 on: Jun 8th, 2003, 1:12pm » |
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 sorry i forgot that
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SWF
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Re: Combinatorial sum
« Reply #4 on: Jun 11th, 2003, 10:11pm » |
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That was a tedious one:
6n*2n + n(13n-24)*2n-1 + n(n-1)(8n-27)*2n-2 + n(n-1)(n-2)(n+6)*2n-3
+ 8n(n-1)(n-2)(n-3)*2n-4 + n(n-1)(n-2)(n-3)(n-4)*2n-5
I worked my way up from 0 to 5 for the exponent on k in the original question by rearranging n!/k!/(n-k)! to (n+1)!/(k+1)!/(n+1-(k+1))!*(k+1)/(n+1)=N+1Ck+1*(k+1)/(N+1) and changing indices on the summation. That leaves an expression that can be used to give the sum for the next higher exponent on k in terms of the lower exponents.
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