A sequence x 0 , x 1 , x 2 , x 3 , …. is defined by letting x 0 = 5 and x k =4+x k–1 for all natural numbers k. Show that x n = 5 for all n ϵ N using mathematical induction.

Question

Philoid · Accepted Answer

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.

A sequence x0, x1, x2, x3, …. is defined by letting x0 = 5 andxk =4+xk–1 for all natural numbers k. Show that xn = 5 for allnϵN using mathematical induction.

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.