Last updated at May 29, 2018 by Teachoo

Transcript

Ex 4.1,23 Prove the following by using the principle of mathematical induction for all n N: 41n 14n is a multiple of 27. Introduction If a number is multiple of 27, then it will come in table of 27 27 1 = 27 27 2 = 54 27 3 = 71 Any number multiple of 27 = 27 Natural number Ex 4.1,23 Prove the following by using the principle of mathematical induction for all n N: 41n 14n is a multiple of 27. Let P(n):41n 14n = 27d , where d N For n=1, L.H.S = 411 141 = 41 14 = 27 = 27 1 = R.H.S P(n) is true for n = 1 Assume P(k) is true 41k 14k = 27m, where m N We will prove that P(k + 1) is true L.H.S = 41k+1 14k+1 = 41k . 411 14k . 141 = 41k . 41 14k . 14 = (27m + 14k) 41 14k . 14 = 41 27m + 41 14k 14k . 14 = 41 27m + 14k (41 14) = 41 27m + 14k (27) = 27 (41m 14k) = 27 r, where r = (41m 14k ) is a 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

Ex 4.1

Ex 4.1, 1
Important

Ex 4.1, 2

Ex 4.1, 3

Ex 4.1, 4

Ex 4.1, 5

Ex 4.1, 6

Ex 4.1, 7 Important

Ex 4.1, 8

Ex 4.1, 9

Ex 4.1, 10

Ex 4.1, 11 Important

Ex 4.1, 12

Ex 4.1, 13 Important

Ex 4.1, 14

Ex 4.1, 15

Ex 4.1, 16

Ex 4.1, 17

Ex 4.1, 18

Ex 4.1, 19

Ex 4.1, 20

Ex 4.1, 21 Important

Ex 4.1, 22

Ex 4.1, 23 Important You are here

Ex 4.1, 24

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.