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, $n=2$
$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$
$S_{n}=S_{2}=1^2+2^2=5$
$S_{4}+4S_{2}=2(2*2+1)^2=50$

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.
