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Remarkably, the sum of the first n cubes is the square of the nth triangular number: 1³ + 2³ + … + n³ = [n(n+1)/2]². So the cubes always add to a perfect square.
Sum of cubes
1³ + 2³ + … + n³ = [n(n + 1) / 2]²
(10 × 11 / 2)² = 55² = 3025.
Because each cube k³ can be split into consecutive odd numbers, and the odd numbers themselves sum to perfect squares.