What is the formula for the sum of cubes of the first n integers?

Σi³ (for i = 1 to n) = [n(n + 1) / 2]². For example, the sum of cubes from 1 to 10 is [10 × 11 / 2]² = 55² = 3,025.

What is Nicomachus's theorem?

It's the classical result stating that the sum of the first n cubes always equals the square of the sum of the first n integers — a surprising connection between two seemingly different sums.

How is this formula proven?

It's typically proven by mathematical induction: verify the base case n = 1, then show that if the formula holds for n, adding (n + 1)³ to both sides still equals [(n + 1)(n + 2) / 2]².

How large a value of n can this calculator handle?

Because the formula is a direct calculation rather than a loop, it computes exact sums for n up to 1 billion instantly, using BigInt arithmetic so the result stays exact no matter how many digits it has.

Is there a similar formula for the sum of squares?

Yes — the sum of the first n squares has its own closed-form formula, n(n + 1)(2n + 1) / 6, covered by the site's dedicated Sum of Squares Calculator.

Does this work for summing cubes starting from a number other than 1?

This calculator computes the sum starting from 1. To find the sum of cubes over a custom range, compute the sum up to the larger endpoint and subtract the sum up to (start − 1).

Why is the sum of cubes always a perfect square?

It follows directly from Nicomachus's theorem: since the sum of cubes equals [n(n + 1)/2]², and n(n + 1)/2 is always a whole number (the nth triangular number), its square is always a perfect square.

Where is the sum of cubes formula used in real life?

It appears in combinatorics, in deriving other polynomial-sum identities, and in verifying closed-form solutions when analyzing the time complexity of triple-nested loop algorithms.

Sum of Cubes Calculator — Add the Cubes of the First n Integers

Our Sum of Cubes Calculator computes 1³ + 2³ + 3³ + ... + n³ using the closed-form formula [n(n + 1) / 2]², giving an exact answer instantly for any n, computed with BigInt precision so even very large sums stay exact rather than approximated.

Quick Answer

The sum of cubes of the first n integers equals [n(n + 1) / 2]² — the square of the sum of the first n integers. Enter n below to instantly get the exact sum, computed with BigInt precision.

Enter n, then click Calculate.

How to Use the Sum of Cubes Calculator — Σi³ Formula Online

1
Enter n, the number of consecutive integers (starting from 1) to cube and sum.
2
Click 'Calculate' to instantly get the exact sum.
3
Review the formula breakdown showing exactly how the sum was computed.

Why Use Sum of Cubes Calculator — Σi³ Formula Online?

The sum of the first n cubes has a remarkably elegant closed-form formula: [n(n + 1) / 2]² — literally the square of the sum of the first n integers. This surprising identity, known as Nicomachus's theorem, means you never need to add cubes one at a time; the calculator computes the exact result in a single step regardless of how large n is, using BigInt arithmetic so precision is never lost even for enormous sums.

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