How to Evaluate a Sum in Sigma Notation

You have a sigma notation sum to evaluate by hand. The Greek capital sigma (Σ) tells you to add a list of terms, each generated by a formula. Leonhard Euler introduced this notation in 1755. To learn how to calculate sigma notation, you read the notation left to right: the index variable, its lower bound, its upper bound, and the expression (the summand) to the right. The index takes every integer from the lower bound to the upper bound, inclusive. You evaluate the expression at each of those integers and add the results. There are two main ways to find the total: expand and add each term, or use summation properties and closed-form formulas. This walks you through both, with worked examples and the mistakes that trip up most students.

This is the direct method. You write out every term listed by the sigma notation, then add them. It works for any sum, no matter how simple or complicated the summand is.

Step 1: Identify the Parts

Find the index variable (often i, j, k, or n), the lower bound (the starting integer), the upper bound (the ending integer), and the summand (the expression to the right of Σ). For example, in Σ_k=2⁵ k², the index is k, the lower bound is 2, the upper bound is 5, and the summand is k².

Step 2: List the Index Values

Write every integer from the lower bound to the upper bound. For bounds 2 to 5, that list is 2, 3, 4, 5. The number of terms is (upper bound - lower bound + 1). So 5 - 2 + 1 = 4 terms.

Step 3: Evaluate the Summand at Each Index Value

Plug each index value into the expression. k=2 gives 2² = 4; k=3 gives 9; k=4 gives 16; k=5 gives 25.

Step 4: Add the Results

Sum the evaluated terms: 4 + 9 + 16 + 25 = 54. That is the total sum.

When the sum has many terms or a pattern, the expand-and-add method becomes slow. Summation properties let you break a complicated sum into simpler parts. The OpenStax Calculus Volume 1 textbook (section 5.1) defines three rules. Use them to evaluate sigma notation without writing every term.

Constant Multiple Rule

If every term has the same constant factor, pull it outside the Σ: Σ(c·aᵢ) = c·Σ aᵢ. For example, Σ_i=1⁴ 3·i = 3·Σ_i=1⁴ i.

Sum/Difference Rule

Splits the sum of two parts into two sums: Σ(aᵢ ± bᵢ) = Σaᵢ ± Σbᵢ. Σ_i=1⁴ (i + 2) = Σ_i=1⁴ i + Σ_i=1⁴ 2.

Splitting Rule

Breaks a sum at an interior bound: Σ_i=mⁿ aᵢ = Σ_i=m^k aᵢ + Σ_i=k+1ⁿ aᵢ.

Closed-Form Formulas

These are shortcuts that give the sum without listing every term. From the same OpenStax section:

• Sum of first n integers: Σ_i=1ⁿ i = n(n+1)/2.
• Sum of first n squares: Σ_i=1ⁿ i² = n(n+1)(2n+1)/6.
• Sum of first n cubes: Σ_i=1ⁿ i³ = [n(n+1)/2]².
• Arithmetic series: Σ_i=1ⁿ (a + (i-1)d) = n/2 · (2a + (n-1)d).
• Finite geometric series: Σ_i=0^n-1 arⁱ = a(1-rⁿ)/(1-r) for r≠1.

The most common error in sigma notation is mis-counting the terms. The number of terms is upper bound - lower bound + 1. If the bounds are 1 to 5, the terms are 1,2,3,4,5: that is 5 - 1 + 1 = 5. If the bounds are 0 to 4, the terms are 0,1,2,3,4: 4 - 0 + 1 = 5 terms. And if the bounds are 3 to 7, it is 7 - 3 + 1 = 5 terms again. The 'plus 1' is what catches you: the lower bound and upper bound are both included. An off-by-one error here throws off every following calculation.

Example 1: Linear Summand, Evaluated Both Ways

Problem: Σ_n=1⁴ (3n - 2).

Expand and add: n=1 gives 3(1)-2 = 1; n=2 gives 4; n=3 gives 7; n=4 gives 10. Sum = 1+4+7+10 = 22.

Properties and formulas: Use the sum/difference rule: Σ(3n - 2) = 3·Σn - Σ2. For bounds 1 to 4: Σn = 4(4+1)/2 = 10. Σ2 = 2+2+2+2 = 8. So 3·10 - 8 = 30 - 8 = 22.

Example 2: Quadratic Summand, Evaluated Both Ways

Problem: Σ_k=2⁵ k².

Expand and add: k=2 gives 4; k=3 gives 9; k=4 gives 16; k=5 gives 25. Sum = 4+9+16+25 = 54.

Properties and formulas: You cannot directly use the formula for Σk² because it starts at 1, not 2. Use the splitting rule: Σ_k=1⁵ k² = Σ_k=1¹ k² + Σ_k=2⁵ k². Or: Σ_k=2⁵ k² = Σ_k=1⁵ k² - Σ_k=1¹ k². Σ_k=1⁵ k² = 5(5+1)(2·5+1)/6 = 55. Σ_k=1¹ k² = 1. So 55 - 1 = 54.

Example 3: Starting at 0

Problem: Σ_i=0³ (2ⁱ).

Expand and add: i=0 gives 2⁰ = 1; i=1 gives 2; i=2 gives 4; i=3 gives 8. Sum = 1+2+4+8 = 15.

Geometric series formula: a=1, r=2, n=4. Sum = 1(1-2⁴)/(1-2) = (1-16)/(-1) = 15.

Example 4: Starting at 3

Problem: Σ_m=3⁶ m.

Expand and add: m=3 gives 3; m=4 gives 4; m=5 gives 5; m=6 gives 6. Sum = 18.

Formula with index shift: This is an arithmetic series. First term = 3, last term = 6, number of terms = 6-3+1 = 4. Sum = 4/2 · (3+6) = 2 · 9 = 18.

Even with the method clear, small errors creep in. Here are the ones that happen most often.

• Off-by-one error in the number of terms. Do not write b - a. Write b - a + 1. For Σ_n=3⁷, the terms are 3,4,5,6,7, which is 5 terms, not 4.
• Off-by-one when applying a closed form. The formula Σi = n(n+1)/2 works for i starting at 1 only. If the lower bound is 3, you must subtract the sum of the first two terms.
• Should you apply the exponent to the constant? In a summand like (2n)², you square 2n, not just n. For n=3, (2·3)² = 36, not 2·9 = 18. In a summand like 2n², you square n first, then multiply by 2. For n=3, 2·9 = 18, not 36. PEMDAS rules.
• Confusing Σx² with (Σx)². Σx² means square each x then add. (Σx)² means add all x first, then square. They are almost never equal. This mistake appears in statistics when computing variance.
• Pulling out a factor that depends on the index. The constant multiple rule only works for a factor that is constant with respect to the index. You cannot pull n out of Σn·c.
• Assuming a sum always has a closed form. Many summands, like Σ sin(i), have no simple formula. The only option is to evaluate term by term.