Unit 9: Sequences, Series, and Combinatorics · ~20 min
Objective
Write and evaluate sums using sigma notation, and apply properties of summation and closed-form formulas to simplify series.
Lessons 9.1 and 9.2 wrote out sums like \(a_{1}\) + \(a_{2}\) + … + \(a_{n}\) in words. Sigma notation (using the Greek capital letter Σ, "sum") compresses that into one compact symbol that names exactly which terms are being added.
\(\sum_{i=1}^{5}\) i² means "add i² for every integer i from 1 to 5" = 1+4+9+16+25 = 55
The letter i here is the index of summation — a placeholder that counts up through every integer from the lower limit to the upper limit, substituting each value into the formula after Σ and adding the results. Any letter can serve as the index (i, k, and n are all common); it disappears once the sum is evaluated.
Properties of summation
Because addition is commutative and associative, sums split apart and constants factor out exactly the way you'd expect:
Constant multiple Σc·\(a_{i}\) = c·Σ\(a_{i}\) — a constant factor can move outside the sum.
Sum/difference Σ(\(a_{i}\) ± \(b_{i}\)) = Σ\(a_{i}\) ± Σ\(b_{i}\) — a sum of sums splits into separate sums.
Constant term \(\sum_{i=1}^{n}\) c = n·c — adding the same constant n times is just n·c.
Closed-form formulas
Some sums come up often enough that it's worth having their totals in closed form, so you never need to add term by term:
Sum
Closed form
\(\sum_{i=1}^{n}\) i
n(n+1)/2
\(\sum_{i=1}^{n}\) i²
n(n+1)(2n+1)/6
\(\sum_{i=1}^{n}\) i³
[n(n+1)/2]²
Notice the first formula is exactly the arithmetic series formula from Lesson 9.1, applied to the sequence 1, 2, 3, …, n (\(a_{1}\) = 1, d = 1). Lesson 9.4 will prove these formulas hold for every positive integer n using mathematical induction.
Worked Example 1 · Expanding and evaluating
ProblemEvaluate \(\sum_{k=1}^{4}\) (2k + 1) by expanding term by term.
1k=1: 3, k=2: 5, k=3: 7, k=4: 9.
23 + 5 + 7 + 9 = 24.
24
Worked Example 2 · Using closed-form formulas and properties
ProblemEvaluate \(\sum_{i=1}^{10}\) (3i² − 4) without expanding all 10 terms.
1Split using summation properties: 3Σi² − Σ4.
2\(\sum_{i=1}^{10}\) i² = 10(11)(21)/6 = 385, so 3(385) = 1155.
3\(\sum_{i=1}^{10}\) 4 = 10(4) = 40.
41155 − 40 = 1115.
1115
Guided practice
1.Write \(\sum_{i=1}^{4}\) 3i in expanded form and evaluate.
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3 + 6 + 9 + 12 = 30.
2.Write the sum 2 + 5 + 8 + 11 + 14 in sigma notation.
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Terms are 3i−1 for i=1…5: \(\sum_{i=1}^{5}\) (3i−1).
3.Evaluate \(\sum_{n=1}^{8}\) n using the closed-form formula.
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n(n+1)/2 = 8(9)/2 = 36.
4.Evaluate \(\sum_{i=1}^{6}\) i³ using the closed-form formula.
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[n(n+1)/2]² = [6(7)/2]² = 21² = 441.
5.Evaluate \(\sum_{i=1}^{5}\) (2i² + 3) using summation properties.
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