# If S1, S2, S3 are the sum of first n natural numbers,

Question:

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 $9 \mathrm{~S}_{2}^{2}=\mathrm{S}_{3}\left(1+8 \mathrm{~S}_{1}\right)$

Solution:

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 $9 S_{2}^{2}=S_{3}\left(1+8 S_{1}\right)$