Sigma Notation Formulas and Properties

The sigma notation formula you need is not the general shape with the Greek letter and the bounds. The one you need is the list of closed-form expressions that let you skip the addition. The sum of the first n integers is n(n + 1)/2. The sum of the first n squares is n(n + 1)(2n + 1)/6. The sum of the first n cubes is [n(n + 1)/2]². These three formulas, combined with the properties of summation, let you simplify almost any polynomial sum you will see in a calculus or statistics course.

OpenStax Calculus Volume 1, section 5.1 Theorem 5.1 lists the four formulas you will use most often. Stewart, Calculus, Appendix E gives the same set. Memorise them or bookmark them. The formulas assume the sum runs from i = 1 to n.

Σ_{i=1}^{n} c = n·c. A constant term added n times is just n times that constant.

Σ_{i=1}^{n} i = n(n + 1)/2. This is the arithmetic series sum for consecutive integers.

Σ_{i=1}^{n} i² = n(n + 1)(2n + 1)/6. The sum of squares formula. Do not confuse this with (Σ i)², which gives a different number.

Σ_{i=1}^{n} i³ = [n(n + 1)/2]². The sum of cubes formula is the square of the sum of integers.

These four are your starting tools. You combine them with the summation rules below to handle expressions like 3i² − 2i + 1.

Two algebraic properties of summation let you break a complicated summand into pieces that match the formulas above. These come from the same theorem in OpenStax Calculus Volume 1, section 5.1.

Constant Multiple Rule

Σ c·aᵢ = c·Σ aᵢ. A factor that does not depend on the index can be pulled out in front of the sigma. If you have Σ 3i², write it as 3·Σ i². Then apply the sum of squares formula. The failure case: do not pull out a factor that contains the index. Σ i·c is not c·Σ i when c is constant; it already is c·Σ i, but if the factor includes i, you cannot factor it.

Sum/Difference Rule

Σ (aᵢ ± bᵢ) = Σ aᵢ ± Σ bᵢ. You split the sum into separate sigma notations. For Σ (3i² − 2i + 1), rewrite it as 3·Σ i² − 2·Σ i + Σ 1. Each piece now matches one of the four formulas.

These two rules are all you need for polynomial sums. They also apply when the bounds are not 1 to n; you just adjust the lower and upper limits on each piece identically.

A sum does not have to start at 1. If the lower bound is m and the upper bound is n, the structure is the same: Σ_{i=m}^{n} aᵢ = aₘ + aₘ₊₁ + ... + aₙ. The number of terms is n − m + 1. The off-by-one error is the most common mistake here: people write n − m terms and miss the last one.

To use the closed-form formulas when m is not 1, shift the index. Define a new variable j = i − m + 1. Then when i = m, j = 1, and when i = n, j = n − m + 1. Rewrite every occurrence of i in the summand in terms of j. For example, Σ_{i=3}^{7} i becomes Σ_{j=1}^{5} (j + 2). Then apply the sum of integers formula: 5·6/2 + 5·2 = 15 + 10 = 25. A quick check: 3 + 4 + 5 + 6 + 7 = 25, so the shift works.

The index variable is a dummy variable; you can rename it without changing the sum. Stewart, Calculus, Appendix E makes this explicit. If you see Σ_{k=2}^{6} k², rename k to i and proceed normally. The dummy variable does not survive the evaluation.

This example uses every tool introduced so far. Evaluate Σ_{i=1}^{n} (3i² − 2i + 1).

Apply the sum/difference rule and the constant multiple rule: Σ (3i² − 2i + 1) = 3·Σ i² − 2·Σ i + Σ 1.

Now substitute the closed forms: 3·[n(n + 1)(2n + 1)/6] − 2·[n(n + 1)/2] + n·1.

Simplify step by step: 3/6 = 1/2, so the first term is n(n + 1)(2n + 1)/2. The second term is −n(n + 1). The third term is n. Write everything over a common denominator of 2: [n(n + 1)(2n + 1) − 2n(n + 1) + 2n] / 2.

Factor n out of the numerator: n[(n + 1)(2n + 1) − 2(n + 1) + 2] / 2. Expand (n + 1)(2n + 1) = 2n² + 3n + 1. Subtract 2(n + 1) = 2n + 2, giving 2n² + 3n + 1 − 2n − 2 = 2n² + n − 1. Add the remaining +2, giving 2n² + n + 1. The final closed form is n(2n² + n + 1)/2.

Check for n = 2: the sum is (3·1² − 2·1 + 1) + (3·2² − 2·2 + 1) = (3 − 2 + 1) + (12 − 4 + 1) = 2 + 9 = 11. The closed form gives 2(2·4 + 2 + 1)/2 = (8 + 2 + 1) = 11. It matches.

The failure case here is forgetting to distribute the constant multiple across every term of the summand. If you treat Σ (3i² − 2i + 1) as 3·Σ i² − 2·Σ i + Σ 1, you are safe. If you try to use the sum of squares formula on 3i² without factoring out the 3, you will misapply the formula.

Sigma notation breaks if you do not know the conventions for unusual bounds. Three cases matter.

Empty Sum

When the upper bound is smaller than the lower bound, the sum is defined as 0. Σ_{i=5}^{3} i² = 0. This is a convention that keeps summation algebra consistent when you split sums at a cut point. OpenStax and Stewart both assume this rule, though it is rarely stated explicitly in introductory sections. If your calculator gives an error for reversed bounds, it is not following the convention; you need to check the manual.

Single Term

When the bounds are equal, the sum is just the summand evaluated at that single index. Σ_{i=4}^{4} i² = 4² = 16. This is consistent with the general form: there is exactly one term.

Non-Integer Bounds

Standard sigma notation requires integer bounds. If you see a sum like Σ_{i=1.5}^{4.3} i, it is not standard. Some contexts treat the lower bound as the smallest integer greater than or equal to it, and the upper bound as the largest integer less than or equal to it, but this is not universal. Avoid it unless your textbook or software defines it explicitly.

The empty sum is the most commonly missed edge case. If you split a sum at a bound and one piece ends up reversed, do not panic; it is 0.