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Help on induction?

  1. Dec 13, 2005 #1
    How do i prove this [tex] \sum_{r=1}^n (r^5)=1/12n^2(n+1)^2(2n^2+2n-1) [/tex]

    when i have already proven this 2.. That [tex] \sum_{r=1}^n (r^3)=1/4n^2(n+1)^2

    [/tex] and [tex] \sum_{r=1}^n (r^3+3r^5)=1/2n^3(n+1)^3 [/tex]

    The problem is that i know how to solve the first 2 but dun know how to solve the big one using the other 2.
     
    Last edited: Dec 13, 2005
  2. jcsd
  3. Dec 14, 2005 #2

    VietDao29

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    Homework Helper

    So you've proven that:
    [tex]\sum_{r = 1} ^ {n} (r ^ 3 + 3r ^ 5) = \frac{1}{2}n ^ {3} (n + 1) ^ {3}[/tex] and [tex]\sum_{r = 1} ^ {n} r ^ 3 = \frac{1}{4}n ^ {2} (n + 1) ^ {2}[/tex].
    To find: [tex]\sum_{r = 1} ^ {n} r ^ 5[/tex]
    You should note that:
    [tex]\sum_{r = 1} ^ {n} (r ^ 3 + 3r ^ 5) \ = \ \sum_{r = 1} ^ {n} r ^ 3 \ + \ \sum_{r = 1} ^ {n} 3r ^ 5 = \sum_{r = 1} ^ {n} r ^ 3 \ + \ 3\sum_{r = 1} ^ {n} r ^ 5[/tex].
    Rearrange it a bit, we have:
    [tex]\sum_{r = 1} ^ {n} r ^ 5 = \frac{1}{3} \times \left( \sum_{r = 1} ^ {n} (r ^ 3 + 3r ^ 5) - \sum_{r = 1} ^ {n} r ^ 3 \right)[/tex].
    Can you go from here?
     
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