## anonymous 3 years ago Is there any formula for this 1^n +2^n +3^n +...+x^n

1. mathslover

why don't you try for a derivation ?

2. 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

3. anonymous

no standard formula

4. anonymous

Cant it be expressed interms of x and n

5. 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 "?

6. 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.. :|

7. 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

8. terenzreignz

Well, except for the ones @amoodarya mentioned, anyway.