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, $n=2$
$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$
$S_{n}=S_{2}=1^2+2^2=5$
$S_{4}+4S_{2}=2(2*2+1)^2=50$
 A: 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.
A: Let $S_n=an^3+bn^2+cn+d$ where $a,b,c,d$ are rational numbers.
So, $S_{2n}+4S_n=n^3 12a+n^2 8b + n 6c+5d$
$\implies n^3 12a+n^2 8b + n 6c+5d= n(2n+1)^2=4n^3+4n^2+n$
Comparing the coefficients of the different powers on $n$,
$12a=4,8b=4,6c=1,d=0$
So, $6S_n=2n^3+3n^2+n=n(n+1)(2n+1)\implies S_n=\frac{n(n+1)(2n+1)}{6}$
Also, $S_n -S_{n-1}=n^2\implies S_n=n^2+S_{n-1}=\sum_{1≤r≤n}r^2+S_0=\sum_{1≤r≤n}r^2$
Observe that, we don't need to know the nature or formula of $S_n$. Solution of any such difference equation of any positive integer degree can be attempted this way.
A: With the additional information you provided about $S_n$ (the sum of squares of the first $n$ integers), there's a neat solution (among other solutions) that uses the perturbation method described in Concrete Mathematics.
Let $C_n$ denote the sum of cubes of the first $n$ natural numbers. Then
\begin{equation}
\begin{split}
C_{n+1} =& C_n + (n+1)^3 = \sum_{k=1}^{n+1}k^3 = \sum_{k=0}^{n}(k+1)^3 = \sum_{k=0}^{n} k^3 + 3k^2 + 3k + 1 \\
=&\sum_{k=0}^{n}k^3 + 3\sum_{k=0}^{n}k^2 + 3\sum_{k=0}^{n}k + \sum_{k=0}^{n}1 = C_n + 3S_n + 3\dfrac{n(n+1)}{2} + (n+1).
\end{split}
\end{equation}
Hence
\begin{equation}
\begin{split}
S_n =& \dfrac{(n+1)^3}{3} - \dfrac{n(n+1)}{2} - \dfrac{n+1}{3} = \dfrac{(n+1)(2(n+1)^2 - 3n - 2)}{6} \\
=& \dfrac{(n+1)(2n^2+n)}{6} = \dfrac{n(n+1)(2n+1)}{6}.
\end{split}
\end{equation}
A: $S_{2n}+4S_{n}=n(2n+1)^2$
$S_{2n}=S_{n}+S_{(n+1,2n)}$ -------(A)
$S_{(n+1,2n)}=(n+1)^2+(n+2)^2+(n+3)^2+\cdots+(2n)^2$
$S_{(n+1,2n)}=(n+1)^2+(n+2)^2+(n+3)^2+\cdots+(n+n)^2$
$S_{(n+1,2n)}=n(n)^2+(1^2+2^2+3^2+\cdots+n^2)+(2n)(1+2+3+\cdots+n)$
$S_{(n+1,2n)}=n^3+S_{n}+(2n)\frac{n(n+1)}{2}$
$S_{(n+1,2n)}=n^3+S_{n}+n^2(n+1)$
$S_{(n+1,2n)}=n^3+S_{n}+n^3+n^2$
$S_{(n+1,2n)}=S_{n}+2n^3+n^2$ -------(B)
we, have from (A) and (B),
$S_{2n}=S_{n}+S_{n}+2n^3+n^2$
$S_{2n}=2S_{n}+2n^3+n^2$
we, now have,
$6S_{n}+2n^3+n^2=n(2n+1)^2$
$6S_{n}+2n^3+n^2=n(4n^2+4n+1)$
$6S_{n}+2n^3+n^2=4n^3+4n^2+n$
$6S_{n}=2n^3+3n^2+n$
$6S_{n}=n(n+1)(2n+1)$
$S_{n}=\frac{n(n+1)(2n+1)}{6}$
A: You could use induction.
Assuming that $$S_{2n}+4S_n=n(2n+1)^2$$ then add terms to both sides so that the left side increments its index:
$$
\begin{align}
&S_{2n}+4S_n+(2n+1)^2+(2n+2)^2+4(n+1)^2\\
&=n(2n+1)^2+(2n+1)^2+(2n+2)^2+4(n+1)^2\\
S_{2(n+1)}+4S_{n+1}&=n(2n+1)^2+(2n+1)^2+(2n+2)^2+4(n+1)^2\\
&=(n+1)(2n+1)^2+4(n+1)^2+4(n+1)^2\\
&=(n+1)(2n+1)^2+8(n+1)^2\\
&=(n+1)[(2n+1)^2+8(n+1)]\\
&=(n+1)[4n^2+4n+1+8n+8]\\
&=(n+1)[4n^2+12n+9]\\
&=(n+1)(2n+3)^2\\
&=(n+1)(2(n+1)+1)^2\\
\end{align}$$
The base case is established in your question.
A: Hint $\rm\quad S_n =\, \sum c_k n^k\ \Rightarrow\ S_{2n} + 4\, S_n\, =\: \sum\ (2^k\!+\!4)\ c_k\ =\ 4\, n^3+4\,n^2 + n,\:$ therefore
$$\rm S_n\, =\ \frac{4}{2^{\color{#C00}3}\!+\!4} n^\color{#C00}3 +\frac{4}{2^\color{#0A0}2\!+\!4}n^{\color{#0A0}2} + \frac{1}{2^\color{brown}1\!+\!4} n^{\color{brown}1}\ =\ \frac{n^3}3+\frac{n^3}2 + \frac{n}6\ =\ \frac{n(n+1)(2n+1)}6$$
