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Math Help - need help w/ this question.

  1. #1
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    need help w/ this question.

    Hi everybody! I have no math background and I need the ans to this question, I'd really appreciate your help.

    Everytime a new person enters a room all the people that are already in the room shake hands with everybody in the room. people come in one by one. if N people entered the room how many hand shakes were made?

    I kinda get how the algorithm should go but I don't know how to write the answer as a function (maybe using the sum char)
    please help?
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  2. #2
    ux0
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    If your saying...

    n=1 \rightarrow  0 handshakes
    n=2 \rightarrow  1 handshakes
    n=3 \rightarrow  3 handshakes
    n=4 \rightarrow  6 handshakes

    Then the number of each handshake when someone enters is...

    n=n \rightarrow \frac{n(n-1)}{2}

    The total number of handshakes is simple, its just the number of handshakes from n=1 \rightarrow n=N summed

    \sum_{n}^{N} \frac{n(n-1)}{2}
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  3. #3
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    thank you thank you thank you!!!
    just one more thing..
    I know that \sum_{i=1}^{n-1} \ i equals (n(n-1))\2

    but how can I figure out what \sum_{n}^{N} \frac{n(n-1)}{2} equals?
    Last edited by arcticm; October 30th 2009 at 03:33 AM.
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  4. #4
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    Good heavens! Every one who was already in the room shakes hands with everyone else again? I think that would creep me out.

    arcticm, \sum \frac{n(n+1)}{2}= \frac{1}{2}\sum n(n+1).

    It's not to difficult (but tedious) to show that the sum of k degree polynomials in n is a k+1 degree polynomial.

    That is, \sum_{j=0}^N n^2+ n= aN^3+ bN^2+ cN+ d for some number, a, b, c, and d.

    Taking n= 0, 0= a(0^3)+ b(0^2)+ c(0)+ d= d so d= 0.

    Taking n= 1, 1^2+1= 2= a(1^3)+ b(1^3)+ c(1)= a+ b+ c so a+ b+ c= 2.

    Taking n= 2, (2^2+ 2)+ (1^2+ 1)= 8= a(2^3)+ b(2^2)+ c(2)= 8a+ 4b+ 2c so 8a+ 4b+ 2c= 8.

    Taking n= 3, (3^2+ 3)+ (2^2+ 2)+ (1^2+ 1)= 20= a(3^3)+ b(3^2)+ c(3)= 27a+ 9b+ 3c so 27a+ 9b+ 3c= 20.
    We know that d= 0 and have 3 equations to solve for a, b, and c.
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