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This Sum I am unable to solve

  1. Jun 30, 2004 #1
    Sum(from i=0 to N) of (i^4/2-3i^2/2+1)
    Can you tell me how to calculate the value of that sum ?
    Thank you,

    - Pattielli
  2. jcsd
  3. Jun 30, 2004 #2

    matt grime

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    split into two smaller sums, factor out the constants and use the formulae for the sums of 4th and second powers that I haven't memorized but which should be googlable and that the constant term at the end contributes N+1 to the sum.
  4. Jun 30, 2004 #3


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    [tex]\sum _{i=0} ^N \frac{1}{2}i^4 - \frac{3}{2}i^2 + 1[/tex]

    [tex]= 0.5\sum i^4 + 1.5 \sum i^2 + N + 1[/tex]*

    You need to find a general closed-form solution to the sum of the first N squares and first N [itex]i^4[/itex] (whatever you'd call them).

    [tex]\sum _{i=0} ^N = \frac{N(N+1)(2N+1)}{6}[/tex]

    For the [itex]i^4[/itex], I don't have the solution memorized, nor can I think of a way right now to solve it.


    Check this link out. It's probably a good thing to learn, and it also gives the formula you need:

    [tex]\sum _{i=0} ^n = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n[/tex]

    *Edited based on matt grime's observation
    Last edited: Jun 30, 2004
  5. Jun 30, 2004 #4

    matt grime

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    there are N+1 terms in the sum so your last term, comig from the constant, is out by 1.
  6. Jun 30, 2004 #5


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    Just to make this clear :

    [tex]\sum _{i=0} ^N i^2= \frac{N(N+1)(2N+1)}{6}[/tex]

    The sum of fourth powers will be a polynomial of order 5. Use the values of the sum for N=1..5 to solve for the 5 coefficients (clearly, there's no constant term).

    Or Google harder.
  7. Jul 24, 2004 #6


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    gokul 43201 gave the quickest solution method, via lagrange interpolation.

    there is another recursive method, for findign formulas for the first n terms of any series of rth powers of integers, described in a footnote to the courant calculus book vol 1, in the first few pages. when i saw that i knew I had found a real math book, as there was more content in that footnote than in all math books i had used previously.
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