Here is a closed form solution to your recurrence relation obtained by Maple,
$ s(n)={n}^{2}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}}s \left( 1 \right) +\frac{{n}^{3}}{3}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}} \left( \left( -1 \right)^{{\frac {\ln \left( n \right) }{\ln \left( 2 \right) }}} \right)^{-1}+\frac{{n}^{2}}{2}{n}^{{\frac {i\pi }{ \ln \left( 2 \right) }}} \left( \left( -1 \right) ^{{\frac {\ln \left( n \right) }{\ln \left( 2 \right) }}} \right) ^{-1}+\frac{1}{6}\,n{n}^ {{\frac {i\pi }{\ln \left( 2 \right) }}} \left( \left( -1 \right) ^{ {\frac {\ln \left( n \right) }{\ln \left( 2 \right) }}} \right) ^{-1 }-{n}^{2}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}} \,$
Here is a more compact form
$ s(n) = \left( {n}^{2}\cos \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) +i{n}^{2}\sin \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) \right) s \left( 1 \right) -{n}^{2}\cos \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) +\frac{{n}^{3}}{3}+\frac{{n}^{2}}{2} +\frac{n}{6}-i{n}^{2}\sin \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) \,.$
where $s(1)$ is your initial condition. If you plug in $s(1)=1$ in the above formula you get the simple formula, just as it has been mentioned in the comments,
$ \frac{n}{6} \left( n+1 \right) \left( 2\,n+1 \right) \,,$
which is equal to $ \sum_{i=1}^{n} i^2 $.
Note
If you are interested only in finding sums of the form $ \sum_{i=1}^{n} i^m \,, m=1,2,3,\dots $, then they are simple techniques to find them. See here.