Sum of Squares Calculator (Σi²)
Enter a positive integer n; the tool applies the closed form Σᵢ₌₁ⁿ i² = n(n+1)(2n+1)/6 to deliver the sum of the first n squared integers instantly — no looping needed. It also reports the arithmetic sum Σi = n(n+1)/2 and the sum of cubes Σi³ = (Σi)² (Nicomachus's theorem), the standard identities behind induction practice, Faulhaber's formulas, variance derivations and series problems.
Enter an integer between 0 and 10⁷.
Sum of squares Σi²
338,350
Arithmetic sum Σi = n(n+1)/2
5,050
Sum of cubes Σi³ = (Σi)²
25,502,500
Formula
Σᵢ₌₁ⁿ i² = n(n+1)(2n+1) / 6 Σᵢ₌₁ⁿ i = n(n+1) / 2 (= Tₙ) Σᵢ₌₁ⁿ i³ = [n(n+1)/2]² = Tₙ² (Nicomachus's theorem)
- · References: OEIS A000330 (square pyramidal numbers); Wikipedia "Square pyramidal number"; Faulhaber's formulas (general Bernoulli-number form).
- · First few Σi²: n=1→1, n=2→5, n=3→14, n=4→30, n=5→55, n=10→385, n=100→338,350.
- · Geometric meaning: stack a square pyramid of cannonballs with n layers — layer k has k² balls, so the total is Σi². Galileo (1638) used this identity to derive the s = ½ g t² law of falling bodies.
- · Statistics use: sample variance σ² = (Σx² − n·x̄²)/(n−1); this identity is preferred over the naive form because it reduces accumulated round-off error.
- · Proof: induction on n, or the telescoping difference (k+1)³ − k³ = 3k² + 3k + 1 summed from 1 to n and solved for Σk².
- · Related identities: 1² + 3² + 5² + … + (2n−1)² = n(2n−1)(2n+1)/3 (odd squares); 2² + 4² + … + (2n)² = 2n(n+1)(2n+1)/3 (even squares).
Frequently asked
Why is the closed form n(n+1)(2n+1)/6 — is there an intuitive proof?
The cleanest proof is telescoping. Take (k+1)³ − k³ = 3k² + 3k + 1 and sum k = 1..n. The left side collapses to (n+1)³ − 1; the right side becomes 3·Σk² + 3·Σk + n. Substituting the known Σk = n(n+1)/2 and solving for Σk² yields n(n+1)(2n+1)/6. A combinatorial route: rewrite i² as 2·C(i,2) + C(i,1) and use the hockey-stick identity Σ C(i, k) = C(n+1, k+1) to collapse each sum in one step. Geometric intuition: stack three "L-shaped" square pyramids together — the total fits a rectangular box with edges n, n+1, 2n+1, giving 6·Σi² = n(n+1)(2n+1) directly.
How does Σi² show up in variance and statistical analysis?
The textbook definition of sample variance is σ² = Σ(xᵢ − x̄)² / (n − 1). Expanding the square gives σ² = (Σxᵢ² − n·x̄²) / (n − 1) — "sum of squares minus the square of the mean." This computational form is numerically more stable than the naive one for wide-range data, so it is the form most statistics texts and library implementations use. When xᵢ are exactly 1, 2, …, n the closed form drops in: x̄ = (n+1)/2 and Σxᵢ² = n(n+1)(2n+1)/6, yielding σ² = (n² − 1)/12 — the discrete-uniform variance used in random-ID, lottery and dice analyses. Linear regression uses the same quantity: Sxx = Σ(xᵢ − x̄)² shows up in the slope estimate β̂ = Sxy ⁄ Sxx. For evenly spaced xᵢ = 1..n the denominator is n(n² − 1)/12 in closed form.
How are square pyramidal numbers related to triangular numbers and sums of cubes?
All three belong to the family of figurate numbers: · Triangular number Tₙ = 1 + 2 + … + n = n(n+1)/2 — dots needed to form a triangle (OEIS A000217). · Square pyramidal number Pₙ = 1² + 2² + … + n² = n(n+1)(2n+1)/6 — cannonballs in a square pyramid (OEIS A000330). · Sum of cubes Σi³ = Tₙ² = [n(n+1)/2]² — the square of the triangular number. This is Nicomachus's theorem (1ˢᵗ century AD). The visual proof tiles four triangular arrays Tₙ into a (2n+1) × n rectangle, revealing the (2n+1) factor. Relation: Pₙ = Tₙ × (2n+1)/3. So the square pyramidal number is the triangular number times a small linear factor, and the sum of cubes is the triangular number squared. All three closed forms share the same n, (n+1), (2n+1) factors — convenient to memorise as a set.
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