A sequence a₁, a₂, a₃ ... is defined by letting a₁ = 3 and a_k = 7a_k–1 for all natural numbers k ≥ 2. Show that an = 3.7^n–1 for all natural numbers.

Given; a₁ = 3 and a_k = 7a_k–1

⇒ a₂ = 7 × a₁ = 7 × 3 = 21 = 3.7^2-1

⇒ a₃ = 7 × a₂ = 7² × a₁ = 49 × 3 = 147 = 3.7^3-1

⇒ a₄ = 7 × a₃ = 7² × a₂ = 7³ × a₁ = 343 × 3 = 1029 = 3.7^4-1

⇒ a_n = 7a_n-1 = 3.7^n-1

Let a_m = 7a_m-1 = 3.7^m-1 be true.

⇒ a_m+1 = 7a_m+1-1 = 7a_m = 7.3.7^m-1 = 3.7^(m+1)-1

⇒ a_m+1 is true when a_m is true.

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