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

amoodaryaBest 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

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

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

amoodaryaBest 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

shubhamsrgBest 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

terenzreignzBest 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

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