Consider once again the AP: 1, 5, 9, 13, 17, … Since every term is 4 more than the previous term, another way of writing this sequence is t₁ = 1, t_n = t_n−1 + 4 for n ≥ 2. Verify for yourself that this recursive rule leads to the terms of the sequence.

Step-by-step solution

Idea: Start from the given first term and apply ‘add 4’ again and again; compare the results with the given terms.

1. t₁ = 1 (given). t₂ = t₁ + 4 = 5; t₃ = t₂ + 4 = 9.1 mark
2. t₄ = 9 + 4 = 13; t₅ = 13 + 4 = 17. These match 1, 5, 9, 13, 17, so the rule is verified. (It also agrees with the explicit rule 4n − 3.)1 mark

Answer to write in the exam

Common mistakes that cost marks

• Adding 4 to the position instead of the previous term: t₃ = 3 + 4 = 7. Add 4 to the previous term, 5.
• Starting from n = 1 in the rule. The rule is used for n ≥ 2; the first term is given separately.
• Thinking a recursive rule and an explicit rule must give different sequences. Both describe the same AP.

How this can come in the exam

Try one yourself

Write the first five terms given by t₁ = 2, t_n = t_n−1 + 6 for n ≥ 2.

More questions like this

• Find recursive rules for the APs in the previous exercises.
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• Can you find the sum of the first ten natural numbers without actually adding all of them?
• Can the same approach be used to find the sum of 1 + 2 + 3 + … + 100?
• If we use the notation S_n to represent the sum of the first n natural numbers, then S_n = n(n + 1)2. Can you use this formula to find S₂₀, S₅₀ or S₁₀₀₀?

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