Question

If S₁, S₂ and S₃ are the sums of the first n natural numbers, their squares, their cubes respectively, then show that \(9S^2_2\)= S₃ (1 + 8S₁)

Answer 1

S₁ = \(\frac{n(n+1)}{2}\), S₂ = \(\frac{n(n+1)(2n+1)}{6}\), S₃ = \(\frac{n^2(n+1)^2}{2}\)

∴ \(9S^2_2\) = \(9\big\{\frac{1}{6}n(n+1)(2n+1)\big\}^2\) = \(9\big\{\frac{1}{36}n^2(n+1)^2(2n+1)^2\big\}\)

= \(\frac{1}{4}\)n² (n+1)²(4n²+4n+1) = \(\frac{1}{4}\)n² (n+1)²(4n (n+1)+1)

= \(\frac{1}{4}\)n² (n+1)² (1 + 8\(\big(\frac{1}{2}n(n+1)\big)\) = S₃ (1+8S₁).