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

Σi² (for i = 1 to n) = n(n + 1)(2n + 1) / 6. For example, the sum of squares from 1 to 10 is 10 × 11 × 21 / 6 = 385.

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, it also holds for n + 1 by adding (n + 1)² to both sides and simplifying.

Is this the same as squaring the sum of the first n integers?

No — squaring the sum (1 + 2 + ... + n)² gives a different, larger result than summing the squares individually (1² + 2² + ... + n²). This calculator computes the latter.

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.

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

This calculator computes the sum starting from 1. To find the sum of squares over a custom range, you can compute the sum up to the larger endpoint and subtract the sum up to (start − 1), the same subtraction technique used in the Sum of Positive Integers Calculator.

Is there a similar formula for the sum of cubes?

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

Why does the formula always divide evenly by 6?

n(n + 1)(2n + 1) is always divisible by 6 because among any three consecutive-ish factors like these, the product is guaranteed to include both a multiple of 2 and a multiple of 3 — a number-theoretic fact that falls out of the induction proof.

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

It appears in statistics (computing variance via sum-of-squared-deviations formulas), physics (moment of inertia calculations), and computer science (analyzing the time complexity of certain nested-loop algorithms).

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

Our Sum of Squares Calculator computes 1² + 2² + 3² + ... + n² using the closed-form formula n(n + 1)(2n + 1) / 6, 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 squares of the first n integers equals n(n + 1)(2n + 1) / 6. Enter n below to instantly get the exact sum, computed with BigInt precision so nothing is lost to floating-point rounding.

Enter n, then click Calculate.

How to Use the Sum of Squares Calculator — Σi² Formula Online

1
Enter n, the number of consecutive integers (starting from 1) to square 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 Squares Calculator — Σi² Formula Online?

Just as the sum of the first n positive integers has a simple closed-form formula (proven by Gauss's pairing trick), the sum of their squares has its own closed-form formula, n(n + 1)(2n + 1) / 6, provable by mathematical induction. This lets you skip adding up potentially millions of squared terms one by one — the calculator computes the exact result in a single step regardless of how large n is, using BigInt arithmetic so precision is never lost to floating-point rounding.

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