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 + 8 S₁) 

Answer 1

S₁, S₂ and S₃ are the sums of first n natural numbers, their squares and their cubes respectively.

∴ S₁ = `sum_("r" = 1)^"n" "r" = ("n"("n" + 1))/2`

S₂ = `sum_("r" = 1)^"n" "r"^2 = ("n"("n" + 1)(2"n" + 1))/6`

S₃ = `sum_("r" = 1)^"n" "r"^3 = ("n"^2("n" + 1)^2)/4`

= S₃ (1+ 8S₁) = `("n"^2("n" + 1)^2)/4[1 + 8*("n"("n" + 1))/2]`

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

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

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

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

= `9*[("n"("n" + 1)(2"n" + 1))/6]^2` = 9S₂²

Hence, 9S₂² = S₃(1 + 8S₁).