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sauravshakya
Group Title
Is there any formula for this
1^n +2^n +3^n +...+x^n
 one year ago
 one year ago
sauravshakya Group Title
Is there any formula for this 1^n +2^n +3^n +...+x^n
 one year ago
 one year ago

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mathslover Group TitleBest ResponseYou've already chosen the best response.0
why don't you try for a derivation ?
 one year ago

amoodarya Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

manishsatywali Group TitleBest ResponseYou've already chosen the best response.0
no standard formula
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Cant it be expressed interms of x and n
 one year ago

amoodarya Group TitleBest ResponseYou've already chosen the best response.0
its not easy to write when n is >4 but look 1+2+3+ ...+x=x(x+1)/2 is that "what you mean "?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.3
Well these is no standard formula, you can only derive that formula for summation(i^n) if you know the value of summation(i^ (n1)) which will require knowledge of summation(i^(n2)) and so on.. :
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
This reminds me of a pseries... \[\large \sum_{k=1}^{x}\frac{1}{k^{p}}\] 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 pseries
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Well, except for the ones @amoodarya mentioned, anyway.
 one year ago
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