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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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Given; a1 = 3 and ak = 7ak–1

⇒ a2 = 7 × a1 = 7 × 3 = 21 = 3.72-1

⇒ a3 = 7 × a2 = 72 × a1 = 49 × 3 = 147 = 3.73-1

⇒ a4 = 7 × a3 = 72 × a2 = 73 × a1 = 343 × 3 = 1029 = 3.74-1

⇒ an = 7an-1 = 3.7n-1

Let am = 7am-1 = 3.7m-1 be true.

⇒ am+1 = 7am+1-1 = 7am = 7.3.7m-1 = 3.7(m+1)-1

⇒ am+1 is true when am is true.

∴ By Mathematical Induction an = 3.7n–1 is true for all natural numbers n.

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