Chapter 2 · Sequences · Concept 5
2-20 Recurrence Relations (1)
A patient takes an antibiotic every 24 hours while their body clears a fixed share of what is already there. Tomorrow's level depends only on today's. Nobody hands you a formula for day 30 — you are handed the loop, and have to turn it into one.
Guiding question
What if a sequence is described by its own rule instead of a formula?
Learning outcome. Solve recurrence relations at different levels in order to deduce the terms of the sequence.
Two things pin a sequence down
A sequence is completely determined — every term of it — as soon as you know:
(a) where it starts
the first term
(b) how to take one step
aₙ₊₁ from aₙ
That second one is called a recurrence relation. Given both, you can walk to any term you like — the only cost is time.
And two shapes you already know
aₙ₊₁ = aₙ + d
adding the same number each time — an arithmetic sequence, common difference d
aₙ₊₁ = r aₙ
multiplying by the same number each time — a geometric sequence, common ratio r
So when a question hands you one of these two, you do not walk anywhere. You recognise it, read off a and d (or r), and write the general term straight down from lesson 2-2 or lesson 2-7.
The rule is written with an n in it, so using it means choosing a value for n. Put n = 1 to get a₂, then n = 2 to get a₃ — and each answer feeds the next line.
Warm Up 1 · a₁ = 2, and aₙ₊₁ = 3aₙ + 2
a₁ = 2 — given, no working needed
the next one uses the one you just found
The line that confuses people.
aₙ₊₁ = 3aₙ + 2 with n = 1 becomes a₂ = 3a₁ + 2. The subscript on the left went up by one because n + 1 = 2. Write that line out before putting any numbers in.
1. a₁ = 2 and aₙ₊₁ = 3aₙ + 2. Find a₂, a₃ and a₄.
- 1
Put n = 1 into the rule and write the line out.
a₂ = 3a₁ + 2The left side became a₂ because 1 + 1 = 2. Do this before touching numbers.
Work with a partner. For question 2, both of you say whether the rule adds or multiplies before either writes anything. That one word decides the whole method.
1. Find a₂, a₃ and a₄.
2. Find the general term.
1. Find a₂, a₃ and a₄.
2. Find the general term.
Solve the recurrence aₙ = 6aₙ₋₁ − 11aₙ₋₂ + 6aₙ₋₃ for n ≥ 3, given a₀ = 0, a₁ = 1 and a₂ = 6.
This is well beyond the exercises — every rule so far looked back one step, and this one looks back three. Read it as a tour of what the subject does next.
A recurrence lets you reach any term you like by walking. So why bother turning it into a formula at all?
Think about being asked for the 500th term.
Using the wrong n.
To get a₃ from aₙ₊₁ = 3aₙ − n you put n = 2, not n = 3. The n is the term you are stepping FROM.
Trying to jump to a₄ without a₂ and a₃.
A recurrence only knows how to take one step. Each term needs the one before it.
Reading aₙ₊₁ = 3aₙ + 2 as geometric.
The stray +2 stops it. Only aₙ₊₁ = r aₙ exactly is geometric — that shape is lesson 2-22.
Losing brackets on a negative term.
2 × (−6) + 3 needs the bracket. Written as 2 × −6 + 3 it is easy to drop the minus.
Writing the general term when only terms were asked for.
Read the question. "Find a₂ through a₄" wants three numbers, not a formula.
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