There are lots of ways of deriving the formula for the sum of the squares of the first positive integers (see the PlanetMath article). One way is to observe that the sum must be a cubic in (this is obviously the case as the third differences of the sequence are constant). Then we can use the first few terms of the sequence to find the coefficients of the cubic.

I believe there is no elementary closed form for this (there is a formula in terms of generalised harmonic numbers or the polylogarithm function but these are not elementary).2- what is the equation for the following sum :

(1^0.25)+(2^0.25)+(3^0.25)+ ... +(n^0.25)

CB