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Summation proof

  1. Jan 30, 2005 #1
    Hey!

    Can someone please give me a hint on this :uhh:

    Prove:
    [tex]
    \sum_{n=1}^n i^4 = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}
    [/tex]

    What i've got so far:

    Let [itex]P(n)[/itex] be the statement:
    [tex]
    \sum_{n=1}^n i^4 = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}
    [/tex]

    Let [itex]n=1[/itex] we get;
    [tex]
    \sum_{n=1}^1 i^4 = \frac{1(1+1)(2(1)+1)(3(1^2) + 31 - 1)}{30}
    = \frac{(2)(3)(5)}{30}
    = 1
    [/tex]
    Which is true.

    Assume [itex]P(n)[/itex] is true [itex]\forall k \ge n, k \in\mathbb{Z}[/itex]
    Let [itex]n=k+1[/itex]

    Then we get:
    [tex]
    P(k+1) = \sum_{n=1}^{k+1} i^4 = \bigg( \sum_{n=1}^{k} i^4 \bigg) + (k+1)^4 = \frac{k(k+1)(2k+1)(3k^2 + 3k - 1)}{30} + (k+1)^4
    = \frac{1}{30} \bigg[ k(k+1)(2k+1)(3k^2+3k-1) + 30(k+1)^4 \bigg]
    [/tex]

    and then i tried
    [tex]
    \frac{1}{30} \bigg[ k(k+1)(2k+1)(3k^2+3k-1) + 30(k+1)^4 \bigg] = \frac{k+1}{30} \bigg[ k(2k+1)(3k^2+3k-1) + 30(k+1)^3 \bigg]
    [/tex]

    A well as heaps of other arrangements :frown: :cry: My algebra sucks. It turns into a giant mess!
    These sorts of things seem to require a lot of intuition. (or, whats that word.... practice?)

    Thank you very much! :redface: :redface: :blushing:
     
  2. jcsd
  3. Jan 30, 2005 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Good! Factoring out (k+1) simplifies it. I'm afraid what you are going to have to do now is actually multiply that out:
    multiply (2k)(2k+1)(3k2+ 3k- 1) multiply 30(k+1)3 and add them
    Now factor that. It may help to know that you WANT (since you have already factored out (k+1) (which corresponds to the "n" in the original formula) (k+2)(2k+3)(3k2+ 9k+ 5) which correspond, respectively, to n+1, 2n+1, and 3n2+ 3n-1 with k+1 in place of n.
     
  4. Jan 30, 2005 #3
    Thanks HallsofIvy :smile:

    Hmm...

    [tex]\frac{1}{30} \bigg[ k(k+1)(2k+1)(3k^2+3k-1) + 30(k+1)^4 \bigg] = \frac{k+1}{30} \bigg[ k(2k+1)(3k^2+3k-1) + 30(k+1)^3 \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ k(6k^3 + 6k^2 - 2k + 3k^2 + 3k - 1) + 30(k+1)^3 \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ k(6k^3+9k^2+k-1) + 30(k+1)^3 \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + 30(k+1)^3 \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + 30(k+1)(k^2+2k+1) \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + 30(k^3+2k^2+k+k^2+2k+1) \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + (30k^3+60k^2+30k+30k^2+60k+30) \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + (30k^3+90k^2+90k+30) \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + (30k^3+90k^2+90k+30) \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ 6k^4+39k^3+90k^2+90k+30 \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ 3k(2k^3+13k^2) + 30(3k^2+3k+1) \bigg][/tex]

    [tex]= \frac{k+1}{30} \bigg[ 3k(2k^3+13k^2) + 30(3k(k+1)+1) \bigg][/tex]

    ack!

    i tried getting it to look something like err at least..

    [tex]\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} = \frac{(2n^2+3n+1)(3n^2+3n-1)}{30}[/tex]

    [tex]= \frac{6n^4+3n^3-2n^2+9n^3+9n^2-3n+3n^2+3n-1}{30}[/tex]

    [tex]= \frac{6n^4+12n^3+10n^2-1}{30}[/tex]

    Which, where [itex] n=k [/itex] :

    [tex]\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} = \frac{(k+1)(k+2)(2k+3)(3k^2+9k+5)}{30}[/tex]

    :shy:
     
  5. Jan 30, 2005 #4

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    It might help to multiply it out before doing anything. In other words, prove the statement in its non-factored form.

    There's also a cheap trick you can use. :biggrin: If, for some reason, you know that the answer is a polynomial of degree 5, then you can just try 6 values of n and check you get the right answer. (given (k+1) input-output pairs, there is exactly one degree k polynomial passing through them)
     
  6. Jan 30, 2005 #5

    learningphysics

    User Avatar
    Homework Helper

    There's a mistake above... it should be 91k^2 and 89k.

    I think Hurkyl's method of multiplying out what you're trying to prove...(k+2)(2k+3) etc... and showing that it's the same as the above polynomial is the best way.

    If you want to factor your above polynomial ... then since you "expect" (k+2), and (2k+3) as factors... plug in -2 into the polynomial above to see that it goes to zero. That proves that k+2 is a factor. Also try pluggin in k=-3/2 proves that 2k+3 is a factor ince 2k+3= 2(k+3/2).

    You can do polynomial division then to get the last factor.
     
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