Question

Answer the following:

Given that t_n+1 = 5t_n − 8, t₁ = 3, prove by method of induction that t_n = 5ⁿ⁻¹ + 2

Answer 1

The statement P(n) has L.H.S. a recurrence relation t_n+1 = 5t_n – 8, t₁ = 3 and R.H.S. a general statement t_n = 5^n–1 + 2

Step I:

To prove P(1) is true.

L.H.S. = 3, R.H.S. = 5^1–1 + 2

= 1 + 2

= 3

So P(1) is true

∴ For n = 2, L.H.S = t₂ = 5t₁ – 8 = 5(3) – 8 = 7

R.H.S. = t₂ = 5^2–1 + 2

= 5 + 7

= 2

∴ P(2) is also true

Step II:

Assume P(k) is true

i.e. t_k+1 = 5t_k – 8 and t_k = 5^k–1 + 2

Step III:

To prove P(k + 1) is true

i.e. to prove t_k+1 = 5^k+1–1 + 2 = 5^k + 2

Since t_k+1 = 5t_k – 8 and t_k = 5^k–1 + 2 ......(from step II)

∴ t_k+1 = 5(5^k–1 + 2) – 8 = 5^k + 2

∴ P(k + 1) is true.

From all the steps above,

P(n) : t_n = 5^n–1 + 2 is true for all

n ∈ N where t_n+1 = 5t_n – 8, t₁ = 3.