<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] $S_{2n}+4S_{n}=n(2n+1)^2$
$S_{2n}=S_{n}+S_{(n+1,2n)}$ -------(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,
$6S_{n}+2n^3+n^2=n(2n+1)^2$
$6S_{n}+2n^3+n^2=n(4n^2+4n+1)$
$6S_{n}+2n^3+n^2=4n^3+4n^2+n$
$6S_{n}=2n^3+3n^2+n$
$6S_{n}=n(n+1)(2n+1)$
$S_{n}=\frac{n(n+1)(2n+1)}{6}$
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