If $\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}$ are the sum of first $n$ natural numbers, their squares and their cubes, respectively, show that $9 \mathrm{~S}_{2}^{2}=\mathrm{S}_{3}\left(1+8 \mathrm{~S}_{1}\right)$
From the given information,
$S_{1}=\frac{n(n+1)}{2}$
$\mathrm{S}_{3}=\frac{\mathrm{n}^{2}(\mathrm{n}+1)^{2}}{4}$
Here, $S_{3}\left(1+8 S_{1}\right)=\frac{n^{2}(n+1)^{2}}{4}\left[1+\frac{8 n(n+1)}{2}\right]$
$=\frac{n^{2}(n+1)^{2}}{4}\left[1+4 n^{2}+4 n\right]$
$=\frac{n^{2}(n+1)^{2}}{4}(2 n+1)^{2}$
$=\frac{[n(n+1)(2 n+1)]^{2}}{4}$ ..(1)
Also, $9 S_{2}^{2}=9 \frac{[n(n+1)(2 n+1)]^{2}}{(6)^{2}}$
$=\frac{9}{36}[n(n+1)(2 n+1)]^{2}$
$=\frac{[n(n+1)(2 n+1)]^{2}}{4}$ $\ldots(2)$
Thus, from (1) and (2), we obtain $9 S_{2}^{2}=S_{3}\left(1+8 S_{1}\right)$
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