# Answer the following: Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2 - Mathematics and Statistics

Sum

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

#### Solution

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

Step I:

To prove P(1) is true.

L.H.S. = 3, R.H.S. = 51–1 + 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 = 52–1 + 2

= 5 + 7

= 2

∴ P(2) is also true

Step II:

Assume P(k) is true

i.e. tk+1 = 5tk – 8 and tk = 5k–1 + 2

Step III:

To prove P(k + 1) is true

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

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

∴ tk+1 = 5(5k–1 + 2) – 8 = 5k + 2

∴ P(k + 1) is true.

From all the steps above,

P(n) : tn = 5n–1 + 2 is true for all

n ∈ N where tn+1 = 5tn – 8, t1 = 3.

Concept: Principle of Mathematical Induction
Is there an error in this question or solution?

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4 | Q II. (2) | Page 85