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Grade 7-12 Math Review
Series and Sigma Notation

Grade 7-12 Math Review · Grade 11-12 · Precalculus: Series and Induction

Series and Sigma Notation

A sequence lists. A series adds. Sigma is shorthand for the adding.

Evaluate the sum

the sum of (2k + 1) for k = 1 to 4

A sequence is a list; a series is its total.

2, 4, 6, 8 a sequence four separate terms

2 + 4 + 6 + 8 a series one number: 20

aₙ vs Sₙ nth term vs sum of n terms different questions

"Find the 10th term" and "find the sum of the first 10 terms" are asked with almost the same words and answered with completely different formulas. Read which one is wanted before starting.

Sigma names the rule, the start and the stop.

The letter k is a placeholder. Writing the same sum with i or n changes nothing, which is why it is called a dummy variable.

Going the other way: find the pattern, then the limits.

2, 4, 8, 16, 32, 64

these are 2^k

first term is 2¹ so k starts at 1

the lower limit

last is 2⁶ so k stops at 6

the upper limit

the sum of 2^k from k = 1 to 6

six terms in all

Count the terms as a check: from k = 1 to k = 6 there are 6 − 1 + 1 = 6 of them. The "+ 1" is easy to forget and it is why a sum from 4 to 10 has seven terms, not six.

A sigma can be split and a constant pulled out.

Σ(aₖ + bₖ) = Σaₖ + Σbₖ split a sum

Σ(c · aₖ) = c · Σaₖ pull the constant out

Σc from 1 to n = nc the constant added n times

Σ(aₖ · bₖ) ≠ Σaₖ · Σbₖ NO product rule: taking aₖ = bₖ = k over just two terms, 1·1 + 2·2 = 5, but (1 + 2) × (1 + 2) = 9

Σc is nc, not c

The sum of 3 from k = 1 to 6 is 3 + 3 + 3 + 3 + 3 + 3 = 18, not 3. A term with no k in it still appears once per value of k. This is exactly the slip that turned 24 into 20 in the opening problem.

Variations: a late start, alternating signs, and a long sum written backwards.

Reading the rule, the start and the stop carefully covers every one of these.

Starting above 1: the sum of 2k for k = 3 to 6

6 + 8 + 10 + 12

k = 3, 4, 5, 6: four terms

= 36

6 − 3 + 1 = 4 terms ✓

or 2(Σk to 6 − Σk to 2) = 2(21 − 3) = 36

the same, by properties

The standard sums start at k = 1, so a late start is the full sum minus the missing front.

Alternating signs: the sum of (−1)^k · k from k = 1 to 6

−1 + 2 − 3 + 4 − 5 + 6

(−1)^k is −1 for odd k, +1 for even k

(−1 + 2) + (−3 + 4) + (−5 + 6)

pair them up

= 1 + 1 + 1 = 3

three pairs

(−1)^k only switches the sign; pairing neighbours makes the total quick.

Backwards and long: 5 + 8 + 11 + … + 32

terms go up by 3, first is 5: rule 3k + 2

k = 1 gives 5

3k + 2 = 32, so k = 10

the stop

the sum of (3k + 2) for k = 1 to 10

ten terms

3(55) + 10(2) = 185

Σk to 10 = 55; the 2 is added ten times

Solving "rule = last term" finds the upper limit; counting the terms by eye on a long sum is where slips happen.

Split it, then use the standard sums.

Σ(4k − 1) = 4Σk − Σ1

split and pull out

Σk from 1 to 6 = 21

n(n + 1)2 = 6(7)2

Σ1 from 1 to 6 = 6

1 added six times

4(21) − 6 = 78

check: 3 + 7 + 11 + 15 + 19 + 23 = 78 ✓

The opening sum is 24

With only six terms, writing them out and adding is faster and safer. The properties earn their keep when the upper limit is 50 or 200 — and the check by hand is always available for small cases.

Remember the flow

Sequence lists, series adds · Read the start, the stop and the rule · Terms = stop − start + 1 · Split sums and pull out constants

A constant inside a sigma is added once for every index value, not once in total.

On the exam

Tests ask to evaluate a short sigma, count its terms, or write a pattern in sigma form, and the trap answer usually adds a constant only once. The fastest route for a few terms is to list them and add; for many, split the sum and use n(n + 1)2 with the constant counted n times. A 10-second check: stop − start + 1 should equal the number of terms you wrote down.

Now you try

Practice problem

Evaluate the sum

the sum of (k² − 1) for k = 1 to 4

Hint: there are four terms, so the −1 is subtracted four times, not once. Writing the four terms out is the quickest check.

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