<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] 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.
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