Question

A sequence b₀, b₁, b₂ ... is defined by letting b₀ = 5 and b_k = 4 + b_{k – 1} for all natural numbers k. Show that b_n = 5 + 4n for all natural number n using mathematical induction.

Answer 1

We have b₀ = 5 and b_k = 4 + b_{k – 1}

⇒ b₀ = 5, b₁ = 4 + b₀ = 4 + 5 = 9

And b₂ = 4 + b₁ = 4 + 9 = 13

Let P(n) : bn = 5 + 4n

Step 1: P(1) : b₁ = 5 + 4 = 9 

⇒ 9 = 9 which is true.

Step 2: P(k): b_k = 5 + 4k

Let it be true ∀ k ∈ N.

Step 3: Given that P(k) = 4 + b_{k – 1}

⇒ P(k + 1) = 4 + b_k+1–1

⇒ P(k + 1) = 4 + bk = 4 + 5 + 4k

⇒ P(k + 1) = 5 + 4(k + 1) which is true for P(k + 1).

Hence, P(k + 1) is true whenever P(k) is true.