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mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0why don't you try for a derivation ?

amoodarya
 2 years ago
Best ResponseYou've already chosen the best response.0yes 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

manishsatywali
 2 years ago
Best ResponseYou've already chosen the best response.0no standard formula

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0Cant it be expressed interms of x and n

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

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.3Well 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.. :

terenzreignz
 2 years ago
Best ResponseYou've already chosen the best response.0This 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

terenzreignz
 2 years ago
Best ResponseYou've already chosen the best response.0Well, except for the ones @amoodarya mentioned, anyway.
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