I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, $n=2$
$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$
$S_{n}=S_{2}=1^2+2^2=5$
$S_{4}+4S_{2}=2(2*2+1)^2=50$

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.
