If S₁, S₂ and S₃ are the sums of first n natural numbers, their squares and their cubes respectively, then S₁^4 S₂^2 - S₂^2 S₃^2 S₁^2 + S₃^2 =

Step-by-step solution

We know the standard formulas: S₁ = n(n+1) 2 S₂ = n(n+1)(2n+1) 6 S₃ = ( n(n+1) 2 )^2 Notice that S₃ = S₁^2 . Substitute S₃ = S₁^2 into the numerator: S₁^4 S₂^2 - S₂^2 (S₁^2)^2 S₁^2 + S₃^2 = S₁^4 S₂^2 - S₂^2 S₁^4 S₁^2 + S₃^2 = 0 S₁^2 + S₃^2 = 0 .