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Sigma-evaluate in closed form

  1. Jul 6, 2006 #1
    I have another question:
    It says: Evaluate in closed form
    n(sigma)i=1 (2^i - i^2)

    (The n is above the sigma and the i=1 is below)
    what i did so far is:
    (sigma)2*2^i-1 (sigma)i^2
    =(2(2^n -1)/2-1) - (n(n+1)(2n+1)/6)

    now do I need to do something else to shorten this down... if so can someone help me with it...
  2. jcsd
  3. Jul 6, 2006 #2


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    Science Advisor

    That's already pretty much simplified. If you had to do anything to that, you might take the 2 inside the numerator of the first fraction.
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