Summation notation and finding sums of arithmetic and geometric series
Reserve & Extensions • K-12
Sigma notation (also called summation notation) is a compact way to write the sum of many terms following a pattern. Instead of writing out dozens of terms, we use the Greek letter sigma -- Σ -- to say "add all these up."
| Component | Name | Role |
|---|---|---|
| Σ | Sigma | Means "sum" |
| k | Index variable | Counter that changes each term |
| 1 | Lower limit | Starting value of k |
| n | Upper limit | Ending value of k |
| ak | General term | Formula evaluated for each k |
Write out and evaluate: Σk=15 (2k + 1)
Substitute k = 1, 2, 3, 4, 5:
(2(1)+1) + (2(2)+1) + (2(3)+1) + (2(4)+1) + (2(5)+1)
= 3 + 5 + 7 + 9 + 11 = 35
To convert a series into sigma notation, identify the pattern in terms of the index variable.
Write in sigma notation: 4 + 8 + 12 + 16 + 20 + 24
Each term is 4k where k goes from 1 to 6.
Σk=16 4k
Alternatively: 4 · Σk=16 k (factoring out the constant).
An arithmetic series has a constant difference between consecutive terms. Its sum is:
where n is the number of terms, a1 is the first term, and an is the last term.
The famous special case: Σk=1n k = n(n+1)/2.
A geometric series has a constant ratio between consecutive terms. Its sum is:
where a1 is the first term and r is the common ratio.
Arithmetic: Find Σk=1100 k.
S = 100(101)/2 = 5050
(This is the sum Gauss famously computed as a child.)
Geometric: Find Σk=05 3(2k).
Here a1 = 3, r = 2, n = 6 terms (k = 0 through 5).
S = 3(1 − 26)/(1 − 2) = 3(1 − 64)/(−1) = 3(63) = 189
Check: 3 + 6 + 12 + 24 + 48 + 96 = 189.
In a telescoping series, most terms cancel when the sum is expanded, leaving only a few terms from the beginning and end.
Example: Σk=1n [1/k − 1/(k+1)] = (1 − 1/2) + (1/2 − 1/3) + ... + (1/n − 1/(n+1)) = 1 − 1/(n+1) = n/(n+1).
Watch the lower limit carefully. Σk=04 has 5 terms (k = 0, 1, 2, 3, 4), not 4. The number of terms is (upper limit − lower limit + 1). Getting this wrong leads to incorrect sums.
Σ(cak) = c · Σak (constants factor out). Σ(ak + bk) = Σak + Σbk (sums split). These let you break complex sums into simpler pieces.
1. Expand and evaluate: Σk=14 k2
12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30.
2. Write in sigma notation: 5 + 10 + 15 + 20 + ... + 100.
Σk=120 5k. There are 20 terms (100/5 = 20). Sum = 20(5 + 100)/2 = 1050.
3. Find the sum: Σk=18 3 · 2k−1
Geometric series: a1 = 3, r = 2, n = 8. S = 3(1 − 256)/(1 − 2) = 3(255) = 765.
4. Evaluate the telescoping sum: Σk=150 [1/k − 1/(k+1)]
Most terms cancel. Sum = 1 − 1/51 = 50/51.
5. Find: Σk=1n (3k − 1). Express your answer in terms of n.
Split: 3Σk − Σ1 = 3 · n(n+1)/2 − n = 3n(n+1)/2 − n = n[3(n+1)/2 − 1] = n(3n+3−2)/2 = n(3n+1)/2.