The sum of series of successive cubes is equal to square of the sum of root numbers.

1³ + 2³ = 1+8 = 9 = 3² = (1+2)²

1³ + 2³  + 3³ = 1+8 +27 = 36 =6² = (1+2 + 3)²

1³ + 2³  + 3³ + 4³= 1+8 +27 + 64 = 100 =10² = (1+2 + 3 + 4)²

1³ + 2³  + 3³ + 4³ …… n³ =  (1+2 + 3 + 4…… n )² = [n(n+1)/2]²

This can be proved b considering this series as an arithmetic series with varying common difference.

If there are two terms 1³ + 2³  = 9 = 1 +(1+d₁ ) = 1 + (1+d) or d = d₁ = 7

With three terms 1³ + 2³  + 3³ = 36 = 3 + d₁ + d₂ = 3 + (1+d) +(1+2d) = 3 + 3d

                               d₁ + d₂ = 33 ; d₂ = 26 and d = 11 = 7 +4

With four terms  1³ + 2³  + 3³ + 4³ = 100 = 4+ d₁ + d₂ + d₃ = 4 + 6d

                                d₁ + d₂ + d₃  = 96 , d₃ = 63 and d = 16 = 7 + 4 +5

The mean d varies as d = 7+  [(n+1)(n+2)/2 – 3x4/2] = 7 + [ (n² +3n +2)/2 -6] = 1 +(n² +3n +2)/2 = (n² +3n +4)/2 where n ≥ 2

S³ = 1³ + 2³  + 3³ + 4³ …… (n-1)³  where  <d> =  (n² +3n +4)/2

 

   = n [ 1 + (n-1)/2 d] = n { 1 + [(n-1)/2][ (n² +3n +4)/2] = [n(n+1)/2]²