Every formula you'll meet in this course — the length of a vector, the dot product of two lists of numbers, one entry of a matrix product — is a sum of several terms that all look alike except for one changing number. Writing that changing number out by hand (a_1 + a_2 + a_3 + a_4 + a_5) works for five terms and breaks down completely for fifty. Sigma notation is the fix: one symbol, \sum, that says "add up a whole family of terms," with an index doing the bookkeeping for which term you're on. This section is entirely about reading and writing that index correctly — nothing here is new mathematics, just a compact way to say "add these up" that the rest of the course will use on every page.
Think of \sum as a for-loop, the way you'd write one in code:
total = 0; for i in range(1, n+1): total += a[i]. The index i is the
loop variable. It counts 1, 2, 3, \ldots, n, and on each pass you plug
the current count into the formula a[i]. Then you add that result to a
running total. Renaming the loop variable from i to k doesn't change
what the loop computes. That's exactly why \sum_i a_i and \sum_k a_k
are the same sum. A sum with two indices, like \sum_k a_{ik}b_{kj}, is a
nested-loop situation, but only the inner loop variable k is actually
looping. Think of i and j as arguments you passed in before the loop
started, so they stay fixed while k runs. So when you read a sum, ask
two questions first. Which letter is doing the counting? And which
letters are just sitting still, handed in from outside? Answer those and
the rest of the symbol is bookkeeping.
Sigma notation is the shorthand symbol \sum for a repeated sum, and index bookkeeping is the discipline of tracking exactly which values the placeholder subscript underneath it runs over. For a list of numbers a_1, a_2, \ldots, a_n, sigma notation packages their sum as \sum_{i=1}^{n} a_i = a_1 + a_2 + a_3 + \cdots + a_n. Three pieces make this well-defined:
- i is the index of summation — a placeholder integer, not a fixed number. It has no meaning outside the sum; \sum_{i=1}^{3} a_i and \sum_{k=1}^{3} a_k are the identical sum, since i and k are both just names for "which term."
- i = 1 (below \sum) is the starting value, and n (above \sum) is the ending value. The sum runs over every integer from the starting value to the ending value, inclusive on both ends.
- a_i — the general term — is a formula in i that produces each term as i counts up. Swap a_i for any expression in i and you get a different sum entirely; \sum_{i=1}^{4} i^2 = 1+4+9+16 = 30 is nothing like \sum_{i=1}^{4} 2i = 2+4+6+8=20, even though both run i from 1 to 4.
Two bookkeeping facts matter more here than in most courses, because linear algebra leans on them constantly. First, the starting index doesn't have to be 1: \sum_{i=0}^{n} or \sum_{i=2}^{n} are both legal and common — always read the bounds off the symbol, never assume 1. Second, a sum can carry two indices at once, one for rows and one for columns: \sum_{k=1}^{n} a_{ik} b_{kj} sums over k only, while i and j sit fixed as labels — that's the exact shape you'll compute for every entry of a matrix product later in this course, so getting comfortable with "one index moves, the others don't" now pays off directly.
Sigma notation earns its keep the moment you have to add up more than a handful of terms, or state a rule that has to work for any number of terms — a length-5 list today, a length-500 list tomorrow, same symbol either way. Watching a running total build up term by term, with the current index highlighted, is the fastest way to see how the index drives the sum:
Watch a sum accumulate, term by term
Each bar is one term a_i of the sequence. Step through i and watch the partial sum \sum_{i=1}^{n} a_i grow underneath — then drag the sliders to change how the terms are built.
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Recipe Write an explicit sum in sigma notation Use when: you have a sum written out term by term and need to compress it into one \sum expression Goal: a single \sum expression — index letter, starting value, ending value, and general term — that expands back out to the exact sum you started with
Quick recipe
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1
Line the terms up and compare them. Everything that looks identical from term to term is fixed; the one piece that's different in each term — almost always a subscript — is what the index will track.
Example
In a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4, every term has the same shape, a_{\square}b_{\square}, with a matching pair of subscripts multiplied together. The letters a and b, and the multiplication between them, never change; only the subscript does.
Why: the index is only allowed to replace the ONE thing that's actually changing — treating a fixed piece as if it changed produces a general term that stops matching the real terms after the first one.
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2
Name the index (commonly i, j, or k), then write a formula in that index that reproduces the changing piece exactly as the index counts up by 1 each term.
Example
The subscript in a_1b_1+a_2b_2+a_3b_3+a_4b_4 goes 1, 2, 3, 4 — counting up by exactly 1 each term — so the index itself, with no adjustment, already is the changing piece: general term a_ib_i.
- the changing piece counts by 1 each term (1,2,3,\ldots) — the index itself is the general term's changing piece, with no formula needed on top of it.
- the changing piece skips or scales (e.g. only odd numbers 1,3,5,\ldots, or even numbers 2,4,6,\ldots) — build a small formula in the index, like 2i-1 or 2i, that hits those exact values as i itself counts 1,2,3,\ldots
Why: whichever case applies, this step is what turns "a pattern I can see" into "a formula I can substitute into," which is the only thing a sigma expression is actually allowed to contain.
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3
The first term you wrote tells you where the index starts; the last term tells you where it stops.
Example
The first term, a_1b_1, has subscript 1; the last term, a_4b_4, has subscript 4 — so i runs from 1 to 4.
Why: a sigma expression with a general term but no stated bounds doesn't tell a reader how many terms to add, or where to stop.
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4
Combine the index, its starting and ending values, and the general term into one \sum expression.
Example
a_1b_1+a_2b_2+a_3b_3+a_4b_4 = \sum_{i=1}^{4} a_ib_iWhy: this is the single compact object the previous three steps existed to build — the index, its range, and the formula driving it, all in one symbol.
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5
Substitute the starting value, then every value up to the ending value, back into the general term, and check that the resulting list of terms matches the original sum exactly, in the same order.
Example
Expanding \sum_{i=1}^{4}a_ib_i back out gives a_1b_1+a_2b_2+a_3b_3+a_4b_4 — exactly the sum this recipe started from, confirming the sigma expression is correct.
Why: a general term and bounds can look plausible and still be wrong — expanding back out is the only step that actually confirms they reproduce every original term, not just the first and last.
Common mistakes
- Treating a fixed letter (the a or b itself) as though it were the piece that changes.
- Off-by-one bounds — starting the index at 0 or 2 when the first term's subscript was actually 1.
- Writing a general term that happens to match the first and last term but not the ones in between, because the pattern was assumed to be linear without checking a middle term.
- Skipping the expand-back-out check and shipping a sigma expression that was never actually verified against the original sum.
Check your understanding (counts toward your progress)
Evaluate \sum_{i=1}^{5} 2 (the general term is the constant 2, not 2i).
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
Every one of the 5 terms equals 2 regardless of i, so the sum is just 2 added 5 times: 2\cdot5 = 10.
Correct answer
Also accepted:
Why
Since the term being summed doesn't depend on i at all, each of the five terms in the sum is the same constant.
Evaluate \sum_{i=1}^{3} 3i, the sum of three times each index value.
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
Substitute i=1,2,3 into 3i to get 3, 6, 9, then add: 3+6+9=18.
Correct answer
Also accepted:
Why
Evaluate \sum_{k=2}^{5} (2k - 3).
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
Terms for k=2,3,4,5: 1, 3, 5, 7. Sum: 1+3+5+7 = 16.
Correct answer
Also accepted:
Why
List out the value of 2k-3 for each k from 2 to 5, then add the four results.
Worked example
Evaluate \sum_{i=1}^{5} (2i - 1)
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1read the bounds first — start at i=1, stop at i=5
Write out what the general term 2i-1 produces as i runs from the starting value 1 to the ending value 5, one term per value of i.
read the bounds first — start at i=1, stop at i=5
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2\sum_{i=1}^{5}(2i-1) = \underbrace{(2\cdot1-1)}_{i=1} + \underbrace{(2\cdot2-1)}_{i=2} + \underbrace{(2\cdot3-1)}_{i=3} + \underbrace{(2\cdot4-1)}_{i=4} + \underbrace{(2\cdot5-1)}_{i=5}substitute the index value into the general term, five times
substitute the index value into the general term, five times
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3= 1 + 3 + 5 + 7 + 9simplify each term
simplify each term
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4
Add the five terms left to right: 1+3=4, 4+5=9, 9+7=16, 16+9=25. \sum_{i=1}^{5}(2i-1) = \boxed{25}
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Worked example
Write a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 in sigma notation
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1the pattern is "matching subscript, multiply, add" — Step 1 of the recipe
This is the recipe above, Steps 1–4, run start to finish. Look at what stays the same across every term and what changes. Every term has the shape a_{\square}b_{\square} — a matching pair of subscripts, multiplied together.
the pattern is "matching subscript, multiply, add" — Step 1 of the recipe
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2name the index i, running 1 to 4
The subscript starts at \textcolor{#4f46e5}{1} in the first term and climbs to \textcolor{#4f46e5}{4} in the last term, increasing by exactly 1 each time — that identifies the index, its starting value, and its ending value. \underbrace{a_1b_1}_{\textcolor{#4f46e5}{1}} + \underbrace{a_2b_2}_{\textcolor{#4f46e5}{2}} + \underbrace{a_3b_3}_{\textcolor{#4f46e5}{3}} + \underbrace{a_4b_4}_{\textcolor{#4f46e5}{4}}
name the index i, running 1 to 4
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3
Replace the changing subscript with the index i and attach the bounds you just read off: start at i=1, end at i=4, general term a_ib_i. a_1b_1+a_2b_2+a_3b_3+a_4b_4 = \boxed{\sum_{i=1}^{4} a_ib_i} This exact sum — matching subscripts, multiply, add — is the formula for the dot product of two length-4 lists of numbers, which you'll meet by name in the next chapter.
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Where this shows up: every core formula in this course
Where this goes next
Linear algebra ahead
Sigma notation isn't a warm-up exercise you finish and set aside — it's the language nearly every formula in this course is written in. The dot product of two vectors \mathbf u = (u_1,\ldots,u_n) and \mathbf v = (v_1,\ldots,v_n) is \sum_{i=1}^{n} u_iv_i; the squared length of a vector is \sum_{i=1}^{n} v_i^2; and once matrices arrive, a single entry of a matrix product AB is \sum_{k=1}^{n} a_{ik}b_{kj} — the two-index pattern from this section, with k looping and i, j held fixed as the row and column you're computing.
None of those formulas is new content to learn later; they're this section's sum, wearing different letters. A student who can already read "which letter is looping, which letters are fixed, where do the bounds start and stop" off \sum_{k=1}^{n} a_{ik}b_{kj} meets matrix multiplication as notation they already know applied to a new picture, instead of two new things to learn at once.
Check your understanding (counts toward your progress)
In \sum_{k=1}^{n} 5k, what role does k play?
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.
k isn't fixed at all — it's exactly the opposite, a placeholder that changes value on every term of the sum.
The sum's total is found by adding up the terms, not read directly off the letter k.
The number of terms is n (from the bounds), not 5 — the 5 is just a coefficient in the general term 5k.
Why
k is the index of summation: it takes every integer value from the starting bound (1) to the ending bound (n), one at a time, generating each term.
Write c_1 + c_2 + c_3 + c_4 + c_5 + c_6 using sigma notation. Enter it in the form sum(c_i, i, 1, 6) style is not needed -- just give the general term as a function of i, e.g. "c_i" (assume bounds 1 to 6).
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.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Every term has the shape c_i with the subscript running 1 through 6, so the sum is \sum_{i=1}^{6} c_i; the general term is simply c_i.
Correct answer
Also accepted:
Why
Look at what changes from term to term -- only the subscript increases by 1 each time, from 1 up to 6.
Practice: Sigma notation, index to expression and back (6 problems · doesn’t affect your progress)
In \sum_{k=1}^{n} 5k, what role does k play? (For reference, the choices given are: The index of summation — a placeholder that counts 1, 2, \ldots, n; A fixed constant equal to n; The final answer of the sum; The number of terms being added, always equal to 5.)
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.
k isn't fixed at all — it's exactly the opposite, a placeholder that changes value on every term of the sum.
The sum's total is found by adding up the terms, not read directly off the letter k.
The number of terms is n (from the bounds), not 5 — the 5 is just a coefficient in the general term 5k.
Why
k is the index of summation: it takes every integer value from the starting bound (1) to the ending bound (n), one at a time, generating each term.
The index of summation — a placeholder that counts 1, 2, \ldots, n
Evaluate \sum_{i=1}^{4} i, adding the index values from start to finish.
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
Write out the terms as i runs from 1 to 4: 1+2+3+4 = 10.
Correct answer
Also accepted:
Why
Evaluate \sum_{k=2}^{5} (2k - 3). (Work through it specifically for \sum_{k=2}^{5} (2k - 3).)
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
Terms for k=2,3,4,5: 1, 3, 5, 7. Sum: 1+3+5+7 = 16.
Correct answer
Also accepted:
Why
List out the value of 2k-3 for each k from 2 to 5, then add the four results.
Which sum equals a_2 + a_3 + a_4 + a_5? (For reference, the choices given are: \sum_{i=2}^{5} a_i; \sum_{i=1}^{5} a_i; \sum_{i=2}^{4} a_i; \sum_{i=0}^{5} a_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.
Starting at i=1 adds an extra term, a_1, that isn't in the list — check the first term written out, not just the count of terms.
Stopping at i=4 leaves out the last term, a_5 — the ending value has to match the last subscript actually written.
Starting at i=0 adds a term, a_0, that was never listed — the starting value must match the first subscript shown, which is 2.
Why
The subscripts run 2, 3, 4, 5, so the index starts at i=2 and ends at i=5: \sum_{i=2}^{5} a_i.
\sum_{i=2}^{5} a_i
\sum_{i=1}^{4} a_i and \sum_{j=1}^{4} a_j are... (For reference, the choices given are: The exact same sum — the index name doesn't change what's being added; Different sums, since i and j are different letters; Equal only if a_i = a_j for every term; Impossible to compare without knowing the values of a_1,\ldots,a_4.)
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.
The index is a placeholder, not a label with meaning of its own — renaming it from i to j doesn't change which terms get added or in what order.
This condition would only make sense if i and j were separate quantities being compared; here they're just two different names for the same loop variable.
No numeric values are needed — the two sums are identical purely because the index is just a placeholder name, for any sequence a_1,\ldots,a_4.
Why
The summation index is a bound variable: its name has no effect on the sum's value. Both expressions add a_1+a_2+a_3+a_4.
Write out what each sum expands to term by term and compare -- the name of the index letter doesn't appear in the expanded sum.
The exact same sum — the index name doesn't change what's being added
A matrix product entry is \sum_{k=1}^{3} p_{2k} q_{k5}. Write out the sum as an explicit expression with no sigma (use * for multiplication and + to add), e.g. "p_21*q_15 + ...". (Work through it specifically for \sum_{k=1}^{3} p_{2k} q_{k5}.)
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.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
With i=2 and j=5 fixed, k runs 1 to 3: p_{21}q_{15} + p_{22}q_{25} + p_{23}q_{35}.
Correct answer
Also accepted:
Why
Write out the sum term by term for k=1, k=2, k=3, substituting the correct subscripts into p_{2k} and q_{k5} each time.
What's next
The next section turns to functions — domain, codomain, and composition — the last piece of notation this chapter needs before the course opens with vectors in \mathbb R^2.