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If $S_1, S_2, S_3$ are the sum of first $n$ natural numbers their squares and their cubes respectively then show that $9 S_2^2=S_3\left(1+8 S_1\right)$
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$S_1$ is the sum of the first $n$ natural numbers
$\therefore \quad S_1=\frac{n(n+1)}{2}$
$S_2$ is the sum of the squares of first $n$ natural numbers.
$\therefore \quad S_2=\frac{n(n+1)(2 n+1)}{6}$
$S_3$ is the sum of the cubes of first $n$ natural numbers.
$\therefore \quad S_3=\left\{\frac{n(n+1)}{2}\right\}^2$
Now, $\quad S_3\left(1+8 S_1\right)=\left\{\frac{n(n+1)}{2}\right\}^2\left\{1+8 \cdot \frac{n(n+1)}{2}\right\}$
$=\left\{\frac{n(n+1)}{2}\right\}^2\{1+4 n(n+1)\}$
$=\left\{\frac{n(n+1)}{2}\right\}^2\left(4 n^2+4 n+1\right)=\left\{\frac{n(n+1)}{2}\right\}^2(2 n+1)^2$
$=\frac{[n(n+1)(2 n+1)]^2}{4}$
$=36 \times \frac{1}{4}\left\{\frac{[n(n+1)(2 n+1)]^2}{6}\right\}$
$=9\left\{\frac{[n(n+1)(2 n+1)]^2}{6}\right\}=9 S_2{ }^2$
$\therefore \quad S_1=\frac{n(n+1)}{2}$
$S_2$ is the sum of the squares of first $n$ natural numbers.
$\therefore \quad S_2=\frac{n(n+1)(2 n+1)}{6}$
$S_3$ is the sum of the cubes of first $n$ natural numbers.
$\therefore \quad S_3=\left\{\frac{n(n+1)}{2}\right\}^2$
Now, $\quad S_3\left(1+8 S_1\right)=\left\{\frac{n(n+1)}{2}\right\}^2\left\{1+8 \cdot \frac{n(n+1)}{2}\right\}$
$=\left\{\frac{n(n+1)}{2}\right\}^2\{1+4 n(n+1)\}$
$=\left\{\frac{n(n+1)}{2}\right\}^2\left(4 n^2+4 n+1\right)=\left\{\frac{n(n+1)}{2}\right\}^2(2 n+1)^2$
$=\frac{[n(n+1)(2 n+1)]^2}{4}$
$=36 \times \frac{1}{4}\left\{\frac{[n(n+1)(2 n+1)]^2}{6}\right\}$
$=9\left\{\frac{[n(n+1)(2 n+1)]^2}{6}\right\}=9 S_2{ }^2$
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