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Sum of the powers of natural numbers

  1. Oct 19, 2012 #1
    Hi everyone. I have learned that:
    1+2+3+...=[itex]\frac{n(n+1)}{2}[/itex]
    12+22+32=[itex]\frac{n(n+1)(2n+1)}{6}[/itex]
    I want to know what the general formula of Ʃna, in which n and a are natural numbers, respect to n and a.
     
  2. jcsd
  3. Oct 19, 2012 #2

    Mute

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    See Faulhaber's formula (and the page about Bernoulli numbers, as they appear in the general formula).
     
  4. Oct 19, 2012 #3

    AlephZero

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    Actually, a simpler generalization of
    1+2+3+ ... = n(n+1)/2
    is
    1.2 + 2.3 + 3.4 + ... = n(n+1)(n+2)/3
    1.2.3 + 2.3.4 + 3.4.5 + ... = n(n+1)(n+2)(n+3)/4
    etc.
     
  5. Oct 20, 2012 #4
    Now I know. Thanks for the answer.
     
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