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Sigma Notation

A free Precalculus lesson from the “Sequences, Series, and Discrete Models” unit, with a worked example and practice problems including step-by-step solutions.

Sigma notation is compact adding language: lower bound, upper bound, and rule for each term. This lesson is part of Precalculus: Advanced Functions, so the emphasis is on interpreting behavior, choosing the right representation, and explaining the result clearly rather than memorizing isolated algebra moves.

What you'll learn

Why it matters: Sequences and series connect repeated patterns, finance, computer loops, and discrete approximations.

Worked example

Problem. Evaluate the sum from k = 1 to 4 of k.

  1. Worked Example: First identify exactly what the question is asking: Evaluate the sum from k = 1 to 4 of k.
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. This is 1 + 2 + ... + 4, an arithmetic series.
  4. Use the closed form N(N + 1)/2 with N = 4: 4(5)/2.
  5. So the sum is 10.
  6. Check the result by substituting or estimating: the response should match 10 and make sense in the original problem.

Answer: 10

Practice problems

1. Evaluate the sum from k = 1 to 4 of k.

Show solution
  1. Warm-up: First identify exactly what the question is asking: Evaluate the sum from k = 1 to 4 of k.
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. This is 1 + 2 + ... + 4, an arithmetic series.
  4. Use the closed form N(N + 1)/2 with N = 4: 4(5)/2.
  5. So the sum is 10.
  6. Check the result by substituting or estimating: the response should match 10 and make sense in the original problem.

Answer: 10

2. Evaluate the sum from k = 1 to 5 of 4.

Show solution
  1. The rule is the constant 4, so every one of the 5 terms equals 4.
  2. Adding a constant 5 times gives N * c = 5 * 4.
  3. So the sum is 20.

Answer: 20

3. Evaluate the sum from k = 1 to 6 of 5k.

Show solution
  1. Factor the constant out: 5 times the sum from k = 1 to 6 of k.
  2. The sum of k is N(N + 1)/2 = 21, so the total is 5 * 21.
  3. So the sum is 105.

Answer: 105

4. Evaluate the sum from k = 1 to 7 of (k + 2).

Show solution
  1. Split it: (sum of k) + (sum of the constant 2).
  2. Sum of k = 28; sum of 2 over 7 terms = 14.
  3. So the total is 28 + 14 = 42.

Answer: 42

5. Expand and add: the sum from k = 1 to 8 of k.

Show solution
  1. Core Practice: First identify exactly what the question is asking: Expand and add: the sum from k = 1 to 8 of k.
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. Substitute k = 1, 2, ..., 8 into the rule.
  4. That is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8.
  5. Adding the terms gives 36.
  6. Check the result by substituting or estimating: the response should match 36 and make sense in the original problem.

Answer: 36

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