anonymous
  • anonymous
Is there any formula for this 1^n +2^n +3^n +...+x^n
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathslover
  • mathslover
why don't you try for a derivation ?
amoodarya
  • amoodarya
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
anonymous
  • anonymous
no standard formula

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anonymous
  • anonymous
Cant it be expressed interms of x and n
amoodarya
  • amoodarya
its not easy to write when n is >4 but look 1+2+3+ ...+x=x(x+1)/2 is that "what you mean "?
shubhamsrg
  • shubhamsrg
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.. :|
terenzreignz
  • terenzreignz
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
terenzreignz
  • terenzreignz
Well, except for the ones @amoodarya mentioned, anyway.

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