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sauravshakya

  • one year ago

Is there any formula for this 1^n +2^n +3^n +...+x^n

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  1. mathslover
    • one year ago
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    why don't you try for a derivation ?

  2. amoodarya
    • one year ago
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    yes but is very complicated ! look these 1+2+3 + ...n=n(n+1)/2 1^2+2^2 +...n^2=n(n+1)(2n+1)/6 1^3+2^3+...+n^3=(n(n+1)/2)^2

  3. manishsatywali
    • one year ago
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    no standard formula

  4. sauravshakya
    • one year ago
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    Cant it be expressed interms of x and n

  5. amoodarya
    • one year ago
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    its not easy to write when n is >4 but look 1+2+3+ ...+x=x(x+1)/2 is that "what you mean "?

  6. shubhamsrg
    • one year ago
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    Well these is no standard formula, you can only derive that formula for summation(i^n) if you know the value of summation(i^ (n-1)) which will require knowledge of summation(i^(n-2)) and so on.. :|

  7. terenzreignz
    • one year ago
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    This reminds me of a p-series... \[\large \sum_{k=1}^{x}\frac{1}{k^{p}}\] And then just set p = -n... But we don't have such an expression for the x'th partial sum as far as I know, for any p-series

  8. terenzreignz
    • one year ago
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    Well, except for the ones @amoodarya mentioned, anyway.

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