# Prove by method of induction, for all n ∈ N: Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1 - Mathematics and Statistics

Sum

Prove by method of induction, for all n ∈ N:

Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1

#### Solution

The statement P(n) has L.H.S. a recurrence relation tn+1 = 5tn + 4, t1 = 4 and R.H.S. a general statement tn = 5n – 1

Step I:

To prove P(1) is true.

L.H.S. = 4, R.H.S. = 51 – 1 = 4

∴ P(1) is true.

For n = 2, L.H.S. = t2 = 5t1 + 4 = 24

R.H.S. = t2 = 52 – 1 = 24

∴ P(2) is also true.

Step II:

Assume P(k) is true.

i.e. tk+1 = 5tk + 4 and tk = 5k – 1

Step III:

To prove P(k + 1) is true.

i.e. to prove tk+1 = 5k+1 – 1

Since tk+1 = 5tk + 4 and tk = 5k – 1 .........(From Step II)

∴ tk+1 = 5(5k – 1) + 4 = 5k+1 – 1

∴ P(k + 1) is true.

From all the steps above,

P(n) : tn = 5n – 1 is true for all

n ∈ N where tn+1 = 5tn + 4 , t1 = 4

Concept: Principle of Mathematical Induction
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 15 | Page 74