Question

A sequence x₀, x₁, x₂, x₃, …. is defined by letting x₀ = 5 and 

x_k =4+x_k–1 for all natural numbers k. Show that x_n = 5 for all 

nϵN using mathematical induction.

Answer 1

Let P(n): x_n =5+4n for all nϵN 

Step1: For n=0, 

P(0):x₀=5+4×0=5 

So, it is true for n=0. 

Step2: Let P(k) be true 

Thus, x_k =5+4k 

Now, we need to show P(k+1) is true whenever P(k) is true. 

P(k+1): 

x_k+1 = 4+ x_k+1-1 

=4+x_k 

=4+5+4k 

=5+4(k+1) 

=RHS 

Thus, P(k+1) is true, so by mathematical induction P(n) is true.