Grade by Grade
C. Discrete Functions · Sequences and Series

Three Ways to Represent a Sequence

C1.4 C1.3 C1.1About 75 minutes
Learning goals
  • I can represent a sequence with a recursion formula, function notation and a formula for the nth term.
  • I can switch between a recursion formula and an explicit formula.
  • I can describe what information each representation shows.
  • I can use differences and ratios to find an explicit formula from the first few terms.

Words to know

explicit formula
Another name for the general term: a formula for tₙ in terms of n only, e.g. tₙ = 2ⁿ.
recursive formula
A formula that gives each term from earlier terms, plus the starting term(s).
first differences
tₙ − tₙ₋₁ for each pair of consecutive terms. Constant first differences mean a linear pattern.
second differences
The differences of the first differences. Constant second differences mean a quadratic pattern.
ratio of consecutive terms
tₙ ÷ tₙ₋₁. A constant ratio means an exponential pattern.

Stairs or elevator?

Suppose you need the 50th term of 2, 4, 8, 16, …. With the rule "double the previous term" you must find all 49 terms before it, like climbing a staircase one step at a time. With the formula tₙ = 2ⁿ you just calculate 2⁵⁰ on a calculator. That is the elevator.

Both descriptions are correct. Each one shows different information. Today you will learn to move between them and to choose the right one for the job.

One sequence, three representations

The curriculum's own example is the sequence 2, 4, 8, 16, 32, 64, …. Here are three algebraic ways to represent it:

  • Recursion formula: t₁ = 2, tₙ = 2tₙ₋₁. It shows the relationship between consecutive terms: each term is double the one before.
  • Function notation: f(n) = 2ⁿ, n ∈ ℕ. It shows the type of function: an exponential function, with a discrete domain.
  • Formula for the nth term: tₙ = 2ⁿ. Like function notation, it lets you calculate any term directly: t₁₀ = 2¹⁰ = 1024.

Function notation and the nth-term formula carry the same information (f(n) = tₙ). The recursion formula is different: it is great for seeing how the sequence grows step by step and for spreadsheets, but slow for finding far-away terms.

2, 4, 8, 16, 32, …the termsRecursion formulat₁ = 2, tₙ = 2tₙ₋₁shows: each term isdouble the one beforeFunction notationf(n) = 2ⁿ, n ∈ ℕshows: an exponentialfunction, discrete domainGeneral termtₙ = 2ⁿshows: any term at once,e.g. t₁₀ = 1024
The sequence 2, 4, 8, 16, 32, … in three algebraic forms. Each form highlights different information.

Finding an explicit formula from the terms

To turn a list of terms (or a recursion formula) into an explicit formula, look for a pattern. Three tools help:

  1. Constant first differences d → linear: tₙ = dn + b. Find b using t₁.
  2. Constant second differences → quadratic: tₙ = an² + bn + c, where a = (second difference) ÷ 2. Subtract an² from each term and find the linear pattern that is left.
  3. Constant ratio r → exponential: each term is r times the one before, so the formula has rⁿ (or rⁿ⁻¹) in it.

If none of these works, look for other patterns: alternating signs, fractions with patterns in the numerator and denominator, or a mix.

Worked example

Example 1: Linear (constant first differences)

Find an explicit formula for t₁ = 5, tₙ = tₙ₋₁ + 3.

  1. Write terms: 5, 8, 11, 14, …. The first difference is always 3.
  2. So tₙ = 3n + b. Use t₁ = 5: 3(1) + b = 5, so b = 2.
  3. tₙ = 3n + 2. Check t₄: 3(4) + 2 = 14 ✓.

tₙ = 3n + 2, or f(n) = 3n + 2

Worked example

Example 2: Quadratic (constant second differences)

Find an explicit formula for 2, 6, 12, 20, 30, ….

  1. First differences: 4, 6, 8, 10. Second differences: 2, 2, 2. Constant second differences, so the pattern is quadratic.
  2. a = 2 ÷ 2 = 1, so tₙ = n² + (something linear).
  3. Subtract n² from each term: 2 − 1 = 1, 6 − 4 = 2, 12 − 9 = 3, 20 − 16 = 4. What is left is just n.
  4. So tₙ = n² + n (= n(n + 1)). Check t₅: 25 + 5 = 30 ✓.

tₙ = n² + n

Worked example

Example 3: Exponential (constant ratio)

Find an explicit formula for 3, 6, 12, 24, ….

  1. Ratios: 6 ÷ 3 = 2, 12 ÷ 6 = 2, 24 ÷ 12 = 2. Constant ratio 2.
  2. Write each term as 3 times a power of 2: t₁ = 3 × 2⁰, t₂ = 3 × 2¹, t₃ = 3 × 2², t₄ = 3 × 2³.
  3. The exponent is always one less than the position: tₙ = 3 × 2ⁿ⁻¹.
  4. Check t₄: 3 × 2³ = 24 ✓. (You will use this pattern a lot in the lesson on geometric sequences.)

tₙ = 3(2)ⁿ⁻¹

Worked example

Example 4: From explicit to recursive

Write a recursion formula for (a) tₙ = 5(3)ⁿ and (b) tₙ = n² + 1.

  1. (a) t₁ = 5(3) = 15. Each time n goes up by 1, we multiply by one more 3, so tₙ = 3tₙ₋₁. t₁ = 15, tₙ = 3tₙ₋₁.
  2. (b) t₁ = 1 + 1 = 2. Find the difference algebraically: tₙ − tₙ₋₁ = n² + 1 − [(n − 1)² + 1] = n² − (n² − 2n + 1) = 2n − 1.
  3. So t₁ = 2, tₙ = tₙ₋₁ + 2n − 1. Check: t₂ = 2 + 3 = 5 = 2² + 1 ✓, t₃ = 5 + 5 = 10 = 3² + 1 ✓.

(a) t₁ = 15, tₙ = 3tₙ₋₁ (b) t₁ = 2, tₙ = tₙ₋₁ + 2n − 1

0123456816243240485664ntₙtₙ = 2ⁿ (exponential)tₙ = n² − n + 2 (quadratic)tₙ = 2n (linear)All three start 2, 4. Two ofthem even share t₃ = 8. But byn = 6 they are 12, 32 and 64.
Linear, quadratic and exponential sequences can begin with the same terms and then pull apart. That is why you should check differences and ratios over several terms before choosing a formula.
RepresentationExample (2, 4, 8, 16, …)Best for
Recursion formulat₁ = 2, tₙ = 2tₙ₋₁seeing how each term relates to the last; spreadsheets
Function notationf(n) = 2ⁿ, n ∈ ℕnaming the type of function and its domain
nth-term formulatₙ = 2ⁿfinding any term quickly; checking whether a number is a term
Table of valuesn: 1, 2, 3, 4 / tₙ: 2, 4, 8, 16spotting differences and ratios
Graphpoints (1, 2), (2, 4), (3, 8), …seeing the shape: linear, curved, increasing or decreasing
Choosing a representation.

Careful: a few terms do not decide everything

The terms 2, 4, 8 could continue 16, 32 (doubling) or 14, 22 (the sequence tₙ = n² − n + 2, whose second differences are constant). A list of terms only fixes a sequence when you also know (or assume) the type of pattern. That is why a question usually says "assuming the pattern continues".

Just for fun

Why was the recursion formula always late for school?

It couldn't get to period 4 without stopping at periods 1, 2 and 3 first.

Hands-on activity

Paper folding: three representations

You need: a sheet of paper and a pencil.

  1. Fold the paper in half. Unfold and count the rectangles. Fold it in half twice (both folds in the same direction or crossing, your choice), unfold and count again. Continue for 3, 4 and 5 folds if you can.
  2. Make a table: n (number of folds) and tₙ (number of rectangles).
  3. Write the sequence three ways: a recursion formula, function notation and an nth-term formula.
  4. Use the nth-term formula to predict the number of rectangles after 10 folds (you cannot actually fold that many times!).

What to look for: 2, 4, 8, 16, 32 rectangles. t₁ = 2, tₙ = 2tₙ₋₁; f(n) = 2ⁿ, n ∈ ℕ; tₙ = 2ⁿ. After 10 folds: 2¹⁰ = 1024 rectangles. Each fold doubles the number of layers, which is why it quickly becomes impossible to fold again.

Your turn

1

Write 3, 9, 27, 81, … as a recursion formula, in function notation and as an nth-term formula.

2

Find an explicit formula for t₁ = 4, tₙ = tₙ₋₁ + 6.

3

Find an explicit formula for 1, 4, 9, 16, 25, ….

4

Find an explicit formula for 5, 10, 20, 40, ….

5

Write a recursion formula for tₙ = 7n − 3.

6

Write a recursion formula for tₙ = 4(5)ⁿ.

7

Which representation would you use to find t₁₀₀ of 1, 3, 9, 27, …? Find it.

More practice: warm-up

1

Constant first differences mean the sequence is ____.

2

Constant second differences mean the sequence is ____.

3

A constant ratio between consecutive terms means the sequence is ____.

4

For f(n) = 10 − 2n, find f(1), f(2) and f(3).

5

Write f(n) = 6n as a recursion formula.

More practice: on your own

1

Find an explicit formula for 3, 8, 15, 24, 35, ….

2

Find an explicit formula for 2, 7, 14, 23, 34, ….

3

Write a recursion formula for tₙ = 2n² .

4

Find an explicit formula for 1/3, 1/9, 1/27, 1/81, ….

5

Describe what you can tell about t₁ = 50, tₙ = tₙ₋₁ − 4 just from its recursion formula, without listing terms.

Challenge

1

Find an explicit formula for the triangular numbers 1, 3, 6, 10, 15, … using second differences.

2

Find an explicit formula for 1, −2, 4, −8, 16, …, and write it as a recursion formula.

3

Find an explicit formula for 2, 5, 11, 23, 47, … (t₁ = 2, tₙ = 2tₙ₋₁ + 1). Hint: add 1 to each term.

Check your understanding

Quiz

1.

Which is an explicit formula for t₁ = 7, tₙ = tₙ₋₁ + 5?

2.

What does a recursion formula show most directly?

3.

The sequence 4, 7, 12, 19, 28, … has second differences of 2. Which formula fits?

4.

Which recursion formula matches tₙ = 6(2)ⁿ?

5.

f(n) = 3ⁿ, n ∈ ℕ. What does the "n ∈ ℕ" tell you?

6.

A sequence starts 2, 4, 8. Which statement is correct?

For parents and teachers

Big idea: Expectation C1.4 asks students to represent a sequence with a recursion formula, function notation or an nth-term formula (the curriculum example is 2, 4, 8, 16, 32, 64, … as t₁ = 2, tₙ = 2tₙ₋₁, as f(n) = 2ⁿ and as tₙ = 2ⁿ) and to describe what each one shows. C1.3 connects the nth-term formula to function notation, and C1.1 continues the idea of a sequence as a discrete function.

How to help: Ask your teen to explain the difference between "the rule to get the next one" and "the rule to jump straight to any one". When they find a formula, ask them to check it on two terms they did not use to find it.

Common difficulties:

  • Using n instead of n − 1 in exponents: for 3, 6, 12, …, tₙ = 3(2)ⁿ⁻¹, not 3(2)ⁿ. Substituting n = 1 catches this.
  • Forgetting to divide the second difference by 2 to get a.
  • Expanding (n − 1)² incorrectly when converting an explicit formula to a recursion. (n − 1)² = n² − 2n + 1.
  • Deciding the pattern from only two or three terms.