A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 +

1.	A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.
Answer» Given; A sequence b₀, b₁, b₂ ... is defined by letting b₀ = 5 and b_k = 4 + b_{k – 1} for all natural numbers k. ⇒ b₁ = 4 + b₀ = 4 + 5 = 9 = 5 + 4.1 ⇒ b₂ = 4 + b₁ = 4 + 9 = 13 = 5 + 4.2 ⇒ b₃ = 4 + b₂ = 4 + 13 = 17 = 5 + 4.3 Let b_m = 4 + b_m-1 = 5 + 4m be true. ⇒ b_m+1 = 4 + b_m+1-1 = 4 + b_m = 4 + 5 + 4m = 5 + 4(m+1) ⇒ b_m+1 is true when b_m is true. ∴ By Mathematical Induction b_n = 5 + 4n is true for all natural numbers n.

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

Answer»

Given; A sequence b₀, b₁, b₂ ... is defined by letting b₀ = 5 and b_k = 4 + b_{k – 1} for all natural numbers k.

⇒ b₁ = 4 + b₀

= 4 + 5 = 9

= 5 + 4.1

⇒ b₂ = 4 + b₁

= 4 + 9

= 13

= 5 + 4.2

⇒ b₃ = 4 + b₂

= 4 + 13

= 17

= 5 + 4.3

Let b_m = 4 + b_m-1 = 5 + 4m be true.

⇒ b_m+1 = 4 + b_m+1-1

= 4 + b_m

= 4 + 5 + 4m

= 5 + 4(m+1)

⇒ b_m+1 is true when b_m is true.

∴ By Mathematical Induction b_n = 5 + 4n is true for all natural numbers n.