Enter a start value (n₁) and an end value (n₂) to compute the sum of consecutive cubes from n₁³ through n₂³. Choose between a range sum, the sum of the first n cubes, or enter custom numbers to cube and add. Results include the total sum, the number of terms, and a step-by-step breakdown of individual cube values.
Sum of Cubes
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Number of Terms
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Largest Cube in Sum
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Average Cube Value
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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.
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.
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.
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.
Using the formula [n(n+1)/2]² with n = 10: [10·11/2]² = 55² = 3025. So 1³ + 2³ + 3³ + ... + 10³ = 3025.
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.
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.
If n₁ > n₂, the range is invalid and produces no terms. The calculator will flag this as an error. Make sure your start value is less than or equal to your end value for a valid range sum.