Is there explicit formula for the expression 1n+2n+…+kn?
I know that for n=1 the explicit formula becomes S=k(k+1)/2 and for n=3 the formula becomes S2. But what about general n?
I know there is a way using the Taylor expansion of f(x)=1/(1−x)=1+x+x2+…, by differentiating it and then multiplying by x and then differentiating again. Repeating this n times, we get
ddx(xddx(…xddxf(x))…)=1+2nxn+3nxn….
Now do the same process but with the function g(x)=xk+1f(x). Then subtract them and we get 1+2nxn+…knxn. Because we have the explicit formulas f(x) and g(x) we can find the explicit formula by this process for arbitrary n. A big problem is that as n grows, it is going take a lot of time finding the explicit formula. My question is therefore: are there other ways?
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