




Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month
Divisible
Example 4 Important Deleted for CBSE Board 2023 Exams
Ex 4.1, 23 Important Deleted for CBSE Board 2023 Exams
Ex 4.1, 22 Deleted for CBSE Board 2023 Exams
Example 6 Important Deleted for CBSE Board 2023 Exams You are here
Ex 4.1, 21 Important Deleted for CBSE Board 2023 Exams
Ex 4.1, 19 Deleted for CBSE Board 2023 Exams
Last updated at Feb. 15, 2020 by Teachoo
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