Question

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

Answer 1

S₁ = 1 + 2 + 3 + ... + n = \[\displaystyle\sum_{r=1}^{n}\frac{n(n+1)}{2}\]

S₂ = 1² + 2² + 3² + ... n² = \[\displaystyle\sum_{r=1}^{n}\frac{n(n + 1)(2n + 1)}{6}\]

S₃ = 1³  + 2³ + 3³ + ... + n3 = `sum_("r" = 1)^"n""r"^3 = ("n"^2 ("n" + 1)^2)/4`

R.H.S. = S₃(1 + 8S₁)

= `("n"^2("n" + 1)^2)/4[1 + 8*("n"("n" + 1))/4]`

= `("n"^2("n" + 1)^2)/4(1 + 4"n"^2 + 4"n")`

= `("n"^2("n" + 1)^2)/4(2"n" + 1)^2`

= `(9."n"^2("n" + 1)^2 (2"n" + 1)^2)/36`

= `9[("n"("n" + 1)(2"n" + 1))/6]^2`

= 9S₂²
= L.H.S.