<s>[INST] 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, $n=2$
$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$
$S_{n}=S_{2}=1^2+2^2=5$
$S_{4}+4S_{2}=2(2*2+1)^2=50$
 [/INST] 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
$$\rm S_n\, =\ \frac{4}{2^{\color{#C00}3}\!+\!4} n^\color{#C00}3 +\frac{4}{2^\color{#0A0}2\!+\!4}n^{\color{#0A0}2} + \frac{1}{2^\color{brown}1\!+\!4} n^{\color{brown}1}\ =\ \frac{n^3}3+\frac{n^3}2 + \frac{n}6\ =\ \frac{n(n+1)(2n+1)}6$$
</s>