A recursive rule builds each term from the ones before it

What to assume they know

• Sequences, terms and subscript notation (Order is the point: a sequence carries position as well as value)
• Explicit rules and how a term is generated from its position (An explicit rule computes any term straight from n)
• Substituting into an expression that contains a product, as in 2u + 3
• Reading an inequality such as n ≥ 2 as a condition on where a rule applies
• Arithmetic with negative numbers, for the exercise that starts at −5

What they should be able to do

• Read a recursive rule as an instruction relating a term to the term before it
• Generate the first several terms of a sequence from a recursive rule and its seed
• Explain why the seed is part of the rule, by producing two sequences that share a step and differ in their first term
• State the condition n ≥ 2 correctly and say what would go wrong without it
• Write a recursive rule for a sequence given in list form
• Handle a rule that reaches back two or three terms, and say how many seeds it needs
• Decide whether a stated number is a term of a recursively defined sequence, and say what makes that harder here than with an explicit rule
• Identify the Virahānka–Fibonacci sequence and place its history

Where it usually goes wrong

• "The recursive rule is the lazy version of the real formula." For the Virahānka sequence the recursion is the definition; the chapter offers no explicit rule for it at all, in this chapter or in the summary. Some patterns are natively step-shaped.
• "A step is a rule." "Add 3 each time" is satisfied by infinitely many sequences. Show the seed-1 and seed-−5 lists side by side until the class volunteers that something is missing.
• **"n ≥ 2 is just notation."** Without it the rule would demand a term before the first one. For a rule reaching back two places the condition is n ≥ 3, and for the three-deep exercise it is n ≥ 4 — the number tracks how far back the step reaches, and so does the number of seeds.
• **"tₙ₋₁ means tₙ minus 1."** It means the previous term. Subtracting one from a term and stepping back one position are different operations, and this is where recursive notation most often breaks down.
• "If a rule is recursive I cannot answer membership questions." You can; it just costs a walk. The chapter's own instruction for the 133 question is to keep computing terms until you have passed the candidate — and passing it is the proof, provided the sequence is increasing.
• "Virahānka–Fibonacci starts 1, 1." This chapter starts it 1, 2. Both conventions exist; the printed one here is 1, 2, and an examination answer should follow the seeds it is given rather than a remembered version.
• "Two different rules cannot give the same sequence." End-of-Chapter problems 14 and 15 exist to break exactly that belief.

Questions to check understanding

• Generate the first five terms from a stated recursive rule and its seed
• Write the recursive rule for a sequence given as a list, seed included
• Convert between the explicit and the recursive description of the same simple sequence
• State how many seeds a given rule requires, and why
• Decide whether a stated value is a term of a recursively defined sequence
• Continue the Virahānka–Fibonacci sequence and identify it by name
• Given a rule stated as tₙ₊₁ in terms of tₙ, restate it in the tₙ and tₙ₋₁ form and confirm the two agree — Exercise Set 8.1 item 5 is written the first way and the section's own examples the second

Examples worth working on the board

Values marked verified are worked out here on the chapter's stated rules; this book prints no answer key.

• The same sequence, twice (§8.3, p. 178). For 1, 4, 7, 10, 13, … the page gives the explicit rule tₙ = 3n − 2 and asks the reader to check it, then builds the recursive description: t₂ = t₁ + 3, t₃ = t₂ + 3, t₄ = t₃ + 3, generalised to tₙ = tₙ₋₁ + 3 for n ≥ 2, with t₁ = 1.
• The seed matters — an added comparison, built from data the chapter supplies. The step "add 3" with seed 1 gives 1, 4, 7, 10, 13; the same step with seed −5, which is Exercise Set 8.1 item 5 (p. 179), gives −5, −2, 1, 4, 7. Verified. Two different sequences, one step. This is the cleanest way to show why the seed is part of the rule and not a preliminary.
• Example 3 (§8.3, p. 178). Seed u₁ = 1 and step uₙ = 2uₙ₋₁ + 3 for n ≥ 2. The page prints the working for the next three: u₂ = 5, u₃ = 13, u₄ = 29, and asks whether 133 is a term, saying only that further terms can be computed to check. Verified: continuing, u₅ = 61, u₆ = 125, u₇ = 253 — the sequence steps straight over 133, so 133 is not a term.
• A closing argument for section 5, offered as enrichment rather than as chapter content. Adding 3 to each term of Example 3 gives 4, 8, 16, 32, 64, … so every term of that sequence is 3 less than a power of 2. Since 133 + 3 = 136 is not a power of 2, no amount of further computing can produce 133. The chapter gives no method for converting a recursion into an explicit rule, so present this as a look ahead, not as the expected Class 9 answer. The same trick settles Exercise Set 8.3 item 3 (p. 193, starred): with seed 2 and step tₙ₊₁ = 3tₙ − 2, every term is one more than a power of 3, and 730 = 729 + 1 puts it in position 7. Verified against the printed rule.
• Example 4 (§8.3, p. 178). Seed s₁ = 3 and step sₙ = sₙ₋₁(sₙ₋₁ − 1) for n ≥ 2. The page prints s₂ = 3 × 2 = 6, s₃ = 6 × 5 = 30, s₄ = 30 × 29 = 870. Verified: the fifth term is 870 × 869 = 756030 — worth computing, because the point of this example is how violently a recursion can accelerate when the step uses the term twice.
• The Virahānka–Fibonacci passage (p. 179). Stated rule: V₁ = 1, V₂ = 2, and Vₙ = Vₙ₋₁ + Vₙ₋₂ for n ≥ 3. The page prints the working V₃ = 2 + 1 = 3, V₄ = 3 + 2 = 5, V₅ = 5 + 3 = 8, and the resulting list 1, 2, 3, 5, 8, 13, 21, 34, …, then asks for the next two. Verified: 55 and 89. Note the seeds: this chapter starts the sequence 1, 2, not 1, 1.
• The history, as printed (p. 179). Virahānka set the sequence down in his 7th-century-CE Vṛttajātisamuchaya, arriving at it through Prakrit metre and poetry. It was studied further by Gopāla around 1135 CE and by Hemachandra around 1150 CE, and later by Fibonacci around 1200 CE. Those four dates are the timeline.
• A three-deep rule (Exercise Set 8.1 item 6, p. 180). Seeds T₁ = 1, T₂ = 2, T₃ = 4, step Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ for n ≥ 4, with T₄ to T₈ asked for. Verified: 7, 13, 24, 44, 81. Three seeds, because the step reaches back three terms — this is the item that makes the seed-counting point on its own.
• Two end-of-chapter recursions worth previewing (p. 195, both starred). Problem 14: P₁ = 1, P₂ = 2, and each later term is one more than the total of all terms before it. Verified: 1, 2, 4, 8, 16, 32, 64, 128 — which collapses to the far simpler rule "double the previous term" from the third term on. Problem 15: W₁ = 1, W₂ = 2, and each later term is two more than the total of all terms up to two places back. Verified: 1, 2, 3, 5, 8, 13, 21, 34 — the Virahānka–Fibonacci sequence again, arrived at from a completely different rule. That pair is the best available evidence that a sequence is not the same thing as the rule that happens to generate it.
• Exercise Set 8.1 item 5's second half (p. 179): is 52 a term of the sequence seeded at −5 with step add 3? Verified: yes, in position 20.

Figures to have open

• A step diagram: terms as boxes in a row with a labelled arrow between each pair, and the seed box drawn differently so it is visibly given rather than computed. Standard schematic; this carries sections 1–3.
• A two-track version of the same diagram, one track seeded 1 and one seeded −5, to make the seed argument visible in one frame. Standard schematic.
• A timeline for the four named mathematicians and their dates as printed on p. 179. Standard schematic; the chapter prints no portrait or figure here.
• §8.3 carries no numbered figure, so nothing needs to come from the textbook for this topic.

Where this sits in the book

• Chapter 8, §8.3 Recursive Rule for a Sequence, pp. 178–179, containing Examples 3 and 4.
• The boxed passage headed Virahānka–Fibonacci sequence, p. 179, which carries the two-term rule, the history and the promise that the sequence returns in later classes. It has no section number of its own; the heading is the locator.
• Exercise Set 8.1, pp. 179–180, items 5 and 6 are the recursive items; items 1–4 belong to §8.2 and An explicit rule computes any term straight from n.
• Deliberate cross-references outside this topic: §8.4 gives the recursive form of an arithmetic progression on p. 183; §8.6 gives it for a geometric progression on p. 187; Exercise Set 8.3 item 3 (p. 193) and End-of-Chapter problems 14 and 15 (p. 195) are recursive, and each of the three is marked with an asterisk in its printed list. End-of-Chapter problem 3 is a different question entirely — it counts three-digit multiples of 7, and belongs to Common difference, and why the nth term is a + (n − 1)d.
• The summary's definition of a recursive formula is on p. 196.