Chapter 2 · Sequences · Concept 5
2-21 Recurrence Relations (2)
Last lesson the rule either added the same number every time or multiplied by the same number every time. Now the amount added CHANGES as you go — and if that sounds familiar, it should. You have met this before under another name.
Guiding question
What if the step size changes as you walk?
Learning outcome. Solve recurrence relations at different levels in order to deduce the terms of the sequence.
When the gap is known
If the rule has the shape
aₙ₊₁ = aₙ + f(n)
then f(n) is the general term of the difference sequence — because rearranging gives aₙ₊₁ − aₙ = f(n), which is exactly what a gap is.
when n ≥ 2
aₙ = a₁ + Σ (k = 1 to n − 1) f(k)
This is lesson 2-18, handed to you.
There, you were given the terms and had to work out the gaps. Here you are given the gaps directly. Everything after that is the same — the same Σ, the same n − 1, the same n = 1 check at the end.
Every rule below writes its own gaps in the orange row. Nothing has to be discovered — you can go straight to adding them up.
Warm Up · a₁ = 1
The rule says the gap from one term to the next is f(n) = 2n − 3. So this is a difference sequence — lesson 2-18, except nobody made you work the gaps out.
a₁ = 1 and aₙ₊₁ = aₙ + 2n − 3. Find the general term.
- 1
Rearrange the rule so the gap is on its own.
aₙ₊₁ − aₙ = 2n − 3Now you can see it: the difference sequence has general term 2n − 3.
Work with a partner. Say out loud which Σ formula the gap needs — the counting numbers, the squares, or a geometric sum. That choice is the only real decision here.
Find the general term.
Parts (f) and (g) subtract. That is not a new case — the gap is simply negative, so f(n) = −3ⁿ and the sum comes out negative. Keep the minus outside the Σ and it looks after itself.
A startup launches with 400,000 LE in reserve, so a₁ = 4 (hundred thousand). Each quarter it burns 0.04n thousand more than the last, giving aₙ₊₁ = aₙ − 0.04n. Find a formula for the reserve at the start of any quarter.
Lesson 2-18 gave you printed terms and asked for the gaps. This lesson gives you the gaps and asks for the terms. Which direction is harder, and why?
Think about what could go wrong in each.
Summing f(n) to n instead of n − 1.
The gaps you walk through to reach term n number n − 1. Summing to n lands you on aₙ₊₁.
Changing only some of the n's in the Σ formula.
Every n in ½n(n+1) becomes (n−1), giving ½(n−1)n. Same warning as lesson 2-18.
Using the wrong first term in a geometric gap.
For f(n) = 3ⁿ the gaps start at 3, not 1. Summing n − 1 of them gives 3(3ⁿ⁻¹ − 1)/2.
Treating a subtraction as a different case.
aₙ₊₁ = aₙ − 3ⁿ just means f(n) = −3ⁿ. Keep the minus outside the Σ.
Skipping the n = 1 check.
The formula was only proved for n ≥ 2. Substituting n = 1 and comparing with a₁ is the last step, every time.
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