"The sum of the first n positive integers" is a mouthful, and it gets worse the moment a proof needs to talk about the sum of the first n squares, or the first n terms of some other pattern entirely. Sigma notation compresses any such sum into one symbol regardless of how many terms it hides — exactly the compression Chapter 5's induction proofs lean on constantly.
Sigma notation is a for-loop that adds instead of printing. "For i from 1 to n, add a_i to a running total" is exactly what the sigma symbol says. It is just written as one symbol instead of three lines of pseudocode. Product notation is the same loop, but it multiplies instead of adding. Its running total starts at 1, not 0. That is also why an empty product equals 1: multiplying by nothing should change nothing.
\displaystyle\sum_{i=1}^{n} a_i means a_1 + a_2 + \cdots + a_n — add the terms a_i as i runs from the lower limit (1 here) to the upper limit (n) in steps of 1. i is the index of summation, a placeholder that disappears once the sum is evaluated; it names no particular number. The identical idea with multiplication instead of addition is product notation: \displaystyle\prod_{i=1}^{n} a_i = a_1 \cdot a_2 \cdots a_n. Both allow any starting index, not just 1, and an empty sum (upper limit below the lower limit) equals 0 while an empty product equals 1 — the identity element for each operation. The index's letter is never meaningful: \sum_{i=1}^n a_i and \sum_{j=1}^n a_j name the identical sum, since i and j are both just placeholders that vanish once every term is added up.
Unfold a sum into its terms
Fill in the missing terms of the expanded sum.
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A running total, term by term
Send successive n through -- f(n) computes the partial sum 1+2+\cdots+n directly, confirming the closed form \frac{n(n+1)}{2} matches expanding the sum by hand.
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Recipe Expand a sum or product written in index notation Use when: you see a sigma or pi symbol with a lower limit, an upper limit, and a formula for the general term Goal: the fully written-out list of terms, ready to add or multiply by hand
Quick recipe
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1
Note the index variable's name, its lower limit, its upper limit, and the formula written after the symbol.
Example
For \sum_{i=1}^{5}(2i-1): the index is i, the lower limit is 1, the upper limit is 5, and the general term is 2i-1.
- the upper limit is at least the lower limit — there's at least one term; continue to Step 2.
- the upper limit is below the lower limit — the sum or product is empty; skip straight to Step 3's empty-case rule.
Why: every later step depends on knowing exactly which values to substitute and which expression to substitute them into.
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2
Plug in every integer from the lower limit to the upper limit, one at a time, keeping each result as its own term.
Example
Substituting i=1,2,3,4,5 into 2i-1 gives 1, 3, 5, 7, 9.
Why: this turns one compact symbol into the actual list of numbers that Step 3 combines.
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3
Combine every term from Step 2 using the operation the symbol calls for.
Example
Adding the terms from Step 2: 1+3+5+7+9=25. For \prod_{i=1}^{4} i, multiply the terms 1,2,3,4 instead: 1\cdot2\cdot3\cdot4=24.
- the symbol is \sum (sigma) — add every term together; an empty sum equals 0.
- the symbol is \prod (pi) — multiply every term together; an empty product equals 1.
⚡ Shortcut
When a sum runs over consecutive integers 1 through n with nothing else attached, skip unfolding entirely: \sum_{i=1}^{n} i = \frac{n(n+1)}{2}, the same formula the widget above checks numerically.
Why: the two notations differ only in which operation combines the unfolded terms, since sigma starts its running total at 0 and pi starts at 1.
Common mistakes
- Multiplying the terms of a sum, or adding the terms of a product, by mixing up which symbol calls for which operation.
- Starting the index one value too early or too late, missing the first or last term.
- Forgetting that an empty sum equals 0 while an empty product equals 1, not the other way around.
- Substituting the index into only part of the general term instead of every occurrence of it.
Worked example
Expand and evaluate a sum
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1
This runs all three steps of the recipe above: identify the index and limits, substitute each value, then add the results. Compute \displaystyle\sum_{i=1}^{5} (2i-1).
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2
Unfold the sum term by term, substituting i=1,2,3,4,5 in turn: (2\cdot1-1) + (2\cdot2-1) + (2\cdot3-1) + (2\cdot4-1) + (2\cdot5-1)
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3
= 1+3+5+7+9
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4\boxed{\sum_{i=1}^{5}(2i-1) = 25}
Worth noticing: this is the sum of the first five odd numbers, and it came out to 5^2 — not a coincidence Chapter 5 will prove holds for every n.
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Worked example
Evaluate a product
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1
Same three steps as the recipe above, with Step 3 multiplying the terms instead of adding them. Compute \displaystyle\prod_{i=1}^{4} i.
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2
Unfold the product term by term: 1 \cdot 2 \cdot 3 \cdot 4.
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3\boxed{\prod_{i=1}^{4} i = 24}
This particular product — every integer from 1 to n multiplied together — is common enough to get its own name and symbol, n! ("n factorial"), which Chapter 10 uses constantly for counting.
Where this goes next
Chapters 5, 9, and 10
Sigma notation isn't a one-chapter tool. Chapter 5's induction proofs are built almost entirely out of manipulating sums like \sum_{i=1}^{n} i into a closed form; Chapter 9's recurrence relations unfold into sums when solved by iteration; and Chapter 10's counting formulas — including the binomial theorem — are stated as sums. Getting comfortable unfolding a sum into its terms now, by hand, is what makes each of those later chapters readable on first sight instead of opaque.
Check your understanding (counts toward your progress)
Compute \sum_{i=1}^{4} i.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
1+2+3+4=10.
Correct answer
Also accepted:
Why
Compute \prod_{i=1}^{3} i.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
1\times2\times3=6.
Correct answer
Also accepted:
Why
What does an empty sum (upper limit below lower limit) equal?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
An empty sum equals 0, the identity element for addition -- adding nothing changes nothing.
What does an empty product equal?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
0 is the empty SUM's value, not the empty product's -- multiplying by 0 would force every product to be 0, which defeats the point of an identity element.
Why
An empty product equals 1, the identity element for multiplication.