# Example 6

Last updated at May 3, 2017 by Teachoo

Last updated at May 3, 2017 by Teachoo

Transcript

Example 6 Prove that 2.7n + 3.5n – 5 is divisible by 24, for all n ∈ N. Introduction If a number is divisible by 24, 48 = 24 × 2 72 = 24 × 3 96 = 24 × 4 Any number divisible by 24 = 24 × Natural number Example 6 Prove that 2.7n + 3.5n – 5 is divisible by 24, for all n ∈ N. Let P(n) : 2.7n + 3.5n – 5 = 24d, when d ∈ N For n = 1, L.H.S = 2.71 + 3.51 – 5 = 2.7 + 3.5 – 5 = 14 + 15 – 5 = 24 = 24 × 1 = R.H.S , ∴ P(n) is true for n = 1 Assume P(k) is true 2.7k + 3.5k – 5 = 24m, when m ∈ N We will prove that P(k + 1) is true. L.H.S = 2.7k+1 + 3.5k+1 – 5 = 2.7k . 71 + 3.5k . 51 – 5 = 7. (2.7k) + 5 . 3.5k – 5 = 7 [24m – 3.5k + 5] + 15.5k –5 = 7 × 24m – (7 × 3). 5k + (7.5) + 15.5k – 5 = 7 × 24m – 21. 5k + 35 + 15.5k – 5 = 7 × 24m – 21. 5k + 15.5k + 35 – 5 = 7 × 24m – 6.5k + 30 = 7 × 24m – 6 (5k – 5) (5k – 5) is a multiple of 4 = 7 × 24m – 6 (4p) = 7 × 24m – 24p = 24 (7m – p) = 24 × r; where r = 7m – p, is some natural number. ∴ P(k + 1) is true whenever P(k) is true. ∴By the principle of mathematical induction, P(n) is true for n, where n is a natural number

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .