What is the sum of cubes formula?
The sum of the first n cubes is given by the closed-form formula: [n(n+1)/2]², which is the square of the nth triangular number. For a range from n₁ to n₂, the sum equals S(n₂) − S(n₁ − 1), where S(k) = [k(k+1)/2]². See also our find Digital Root with Digital Root Calculator.
How do you calculate the sum of cubes from n₁ to n₂?
Use the formula: Σk³ (from k=n₁ to n₂) = [n₂(n₂+1)/2]² − [(n₁−1)n₁/2]². For example, the sum of cubes from 2 to 4 is (4·5/2)² − (1·2/2)² = 100 − 1 = 99, which equals 8 + 27 + 64 = 99.
Why is the sum of cubes equal to a perfect square?
The sum of the first n cubes always equals [n(n+1)/2]², which is the square of the nth triangular number. This elegant identity, known since antiquity, means the sum of the first n cubes is always a perfect square when starting from 1.
What is the relationship between sum of cubes and triangular numbers?
The nth triangular number is Tₙ = n(n+1)/2. The sum of the first n cubes equals Tₙ² — the square of the nth triangular number. This relationship ties together two classic sequences in number theory.
What is the sum of the first 10 cubes?
Using the formula [n(n+1)/2]² with n = 10: [10·11/2]² = 55² = 3025. So 1³ + 2³ + 3³ + ... + 10³ = 3025.
How is the sum of cubes formula used in calculus?
In calculus, the sum of cubes formula is used to evaluate Riemann sums for the integral of x³, confirming that ∫₀ⁿ x³ dx = n⁴/4. It helps bridge discrete summation and continuous integration in introductory analysis courses.
Can the start value n₁ be zero or negative?
Zero is allowed and contributes 0³ = 0 to the sum. This calculator supports non-negative integers for n₁ and n₂. For negative integers, use the custom numbers mode to enter any values you wish to cube and sum.