What is sigma notation?
Long sums are tiring to write. Sigma notation is a short way to write them.
∑k=1n ak = a1 + a2 + ... + an
Read it like this:
- ∑ (sigma) means "add up".
- k is the index. It is a counter. It takes the values 1, 2, 3 ... one by one.
- The number below ∑ is where k starts. The number above is where k stops.
- The rule ak on the right tells you what to add for each k.
Example: ∑k=14 (2k + 1) = 3 + 5 + 7 + 9 = 24.
The index letter does not matter. ∑ k and ∑ j give the same answer. The number of terms is (stop − start + 1).
Properties of sigma (the rules)
These four rules save a lot of work.
- Constant out: ∑ c·ak = c · ∑ ak. A number that does not change with k can be taken out.
- Split a sum: ∑ (ak + bk) = ∑ ak + ∑ bk. The same works for minus.
- Adding a constant n times: ∑k=1n c = n·c. (Each term is c, and there are n of them.)
- Change the start: ∑k=mn ak = ∑k=1n ak − ∑k=1m−1 ak.
Careful: ∑ akbk is NOT (∑ ak)(∑ bk), and ∑ ak² is NOT (∑ ak)².
Sums of k, k² and k³: the ready formulas
Three sums come up again and again. Learn them once.
- ∑k=1n k = n(n+1)/2
- ∑k=1n k² = n(n+1)(2n+1)/6
- ∑k=1n k³ = [n(n+1)/2]²
Why the first one works (3D, step 5): write the staircase 1, 2, ..., n. Turn a copy upside down and put it next to the first. Every column now has height n + 1, and there are n columns. So two copies have total n(n+1), and one copy has half of that.
Notice that ∑k³ is the square of ∑k. With these three formulas and the rules above, you can sum any polynomial in k. For example ∑(3k² − 2k + 5) = 3∑k² − 2∑k + 5n.
Sums of various sequences and telescoping
Some sums are not polynomials. A common trick is telescoping: write each term as a difference, so that most terms cancel, like a telescope folding up.
Take 1/(k(k+1)). It can be split: 1/(k(k+1)) = 1/k − 1/(k+1). So
∑k=1n 1/(k(k+1)) = (1 − 1/2) + (1/2 − 1/3) + ... + (1/n − 1/(n+1)) = 1 − 1/(n+1) = n/(n+1).
Only the first and the last pieces survive. Try the 4th button in the 3D free play to see the bars shrink and the sum creep towards 1.
Other useful sums: for an arithmetic sequence, the sum is n/2 × (first + last); for a geometric sequence with ratio r, the sum is a(rⁿ − 1)/(r − 1). Sums with a recurring pattern such as 1·2 + 2·3 + 3·4 + ... are written as ∑ k(k+1) and solved by splitting into ∑k² + ∑k.
Try it: build the sum yourself
Try it at home: take 10 coins. Make a staircase: 1 coin, then 2, then 3, then 4. How many coins? Now make a second staircase upside down and join them. Count the rectangle: 4 × 5 = 20, so one staircase has 10. That is ∑k for n = 4.
Try it in 3D: before you move the slider, guess the sum for n = 8 using the formula. Then check the readout.
Key formulas and definitions
- ∑(k=1..n) a_k = a_1 + a_2 + ... + a_n
- ∑ c·a_k = c·∑ a_k; ∑ (a_k ± b_k) = ∑ a_k ± ∑ b_k
- ∑(k=1..n) c = n·c
- ∑(k=1..n) k = n(n+1)/2
- ∑(k=1..n) k² = n(n+1)(2n+1)/6
- ∑(k=1..n) k³ = [n(n+1)/2]²
- Telescoping: ∑(k=1..n) 1/(k(k+1)) = 1 − 1/(n+1)
Worked examples
1. Write 2 + 4 + 6 + 8 + 10 using sigma notation.
The k-th term is 2k, and k goes from 1 to 5. So the sum is ∑ (k = 1 to 5) 2k.
2. Find ∑ (k = 1 to 4) (3k − 1).
Put k = 1, 2, 3, 4: 2 + 5 + 8 + 11 = 26.
3. Find 1 + 2 + 3 + ... + 100.
∑k = n(n+1)/2 with n = 100: 100 × 101 / 2 = 5050.
4. Find ∑ (k = 1 to 10) (2k + 3).
Split: 2∑k + ∑3 = 2 × 55 + 10 × 3 = 110 + 30 = 140.
5. Find ∑ (k = 3 to 8) k.
Change the start: ∑(1 to 8) k − ∑(1 to 2) k = 36 − 3 = 33. Check: 3+4+5+6+7+8 = 33.
6. Find 1² + 2² + ... + 6².
n(n+1)(2n+1)/6 with n = 6: 6 × 7 × 13 / 6 = 91.
7. Find ∑ (k = 1 to 9) 1/(k(k+1)).
Telescoping: 1/(k(k+1)) = 1/k − 1/(k+1). The sum is 1 − 1/10 = 9/10.
8. Find 1³ + 2³ + 3³ + 4³ + 5³.
[n(n+1)/2]² with n = 5: (15)² = 225.
Common mistakes
- Forgetting how many terms there are: ∑ from k = 3 to 9 has 9 − 3 + 1 = 7 terms, not 6.
- Writing ∑ c = c instead of n·c. The constant is added n times.
- Thinking ∑ a_k² = (∑ a_k)². Try 1² + 2² = 5, but (1 + 2)² = 9.
- Using the formula n(n+1)/2 when the sum does not start at 1. Subtract the missing front part first.