Question

If S₁, S₂ and S₃ are the sum of first n natural numbers, their squares and their cubes respectively then show that 9S₂² = S₃(1+8S₁).

Answer 1

$S_1$ is the sum of first n natural numbers.
e.g., $S_1=1+2+3+4+.....+n=\frac{n(n+1)}{2}$

$S_2$ is the sum of square of first n natural numbers.
e.g., $S_2=1^2+2^2+3^2+.....n^2=\frac{n(n+1)(2n+1)}{6}$

$S_3$ is the sum of cubes of first n natural numbers.
e.g., $S_3=1^3+2^3+3^3+.....+n^3=\left(\frac{n(n+1)}{2}\right)^2$

now, LHS = $9S_2^2$

= $9\frac{n^2(n+1)^2(2n+1)^2}{36}$

= $\frac{n^2(n+1)^2(2n+1)^2}{4}$

RHS = $S_3(1+8S_2)$

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

= $\frac{n^2(n+1)^2}{4}(1+4n^2+4n)$

=$\frac{n^2(n+1)^2(2n+1)^2}{4}$

hence, LHS = RHS