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Question
A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.
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Solution
Given that: a1 = 3
a2 = 7a2 – 1 = 7.a1 = 7.3 = 21
a3 = 7.a3 – 1 = 7.a2 = 7.21 = 147
Let P(n): an = 3.7n – 1, ∀ n ∈ N
Step 1: P(2) : a2 = 3.72 – 1 = 21
⇒ 21 = 21 which is true for P(2).
Step 2: P(k): ak = 3.7k – 1.
Let it be true.
Step 3: ak = 7ak – 1 .......(Given)
Put k = k + 1
ak + 1 = 7ak = 7(3.7k – 1)
= 3.7k+1–1
= `3.7^(("k"+1) – 1`
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.
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