Question 12 Exercise 7.1

Solutions of Question 12 of Exercise 7.1 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 12(i)

Show by mathematical induction that 52n−124 is an integer. Solution: 1. For n=1, then 52n−124=52.1−124=2424=1∈Z Thus it is true for n=1

2. Let it be true for n=k>1 then 52k−124∈Z 3. For n=k+1 then consider 52(k+1)−124=52k+2−124=52k+2−124=52k⋅25−124=(24+1)52k−124=24.52k+52k−124=24.52k24+52k−124=52k+52k−124 Clearly 52k∈Z∀k∈N,

52k−124∈Z by (a).

Thus the given statement is true for n=k+1. Hence by mathematical induction it is true for all n∈N.

Question 12(ii)

Show by mathematical induction that 10n+1−9n−1081 is an integer.

Solution

1. For n=1 then 10n+1−9n−1081=10i+1−9.1−1081=100−9−1081=1∈Z Thus it is true for n=1.

2. Let it be true for n=k, then 10k+1−9k−1081∈Z⋯(i) 3. For n=k+1, then we have 10k+1+1−9(k+1)−1081=10k+1(10)−9k−9−1081=10k+1(9+1)−9k−9−1081=9.10k+1+10k+1−9k−9−1081=9.10k+1−9+10k+1−9k−1081=9(10k+1−1)+9+10k+1−9k−1081=9(10k+1−1)81+10k+1−9k−1081=10k+1−19+10k+1−9k−1081 Now 10k+1−19∈Z∀k∈N and 10k+1−9k−1081∈Z by (i),

hence it is true for n=k+1. Thus by mathematical induction it is true for all n∈N.

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