Question

If the sum of the first $n$ natural numbers is $S_1$ and that of their squares is $S_2$ and cubes is $S_3$, then show that $9S_2^2=S_3(1+8S_1)$.

Answer 1

Sum of first $n$ natural numbers is $\frac{n(n+1)}{2}$

Sum of squares of first $n$ natural numbers is $\frac{n(n+1)(2n+1)}{6}$

Sum of cubes of first $n$ natural numbers is ${\left(\frac{n(n+1)}{2}\right)}^2$

RHS:

$=S_3(1+8S_1)$

$={\left(\frac{n(n+1)}{2}\right)}^2\times (1+8\times \left(\frac{n(n+1)}{2}\right))={\left(\frac{n(n+1)}{2}\right)}^2\times (4n^2+4n+1)={\left(\frac{n(n+1)}{2}\right)}^2\times {(2n+1)}^2={\left(\frac{n(n+1)(2n+1)}{2}\right)}^2=9\times {\left(\frac{n(n+1)(2n+1)}{6}\right)}^2=9{S_2}^2$

Hence proved