Chapter 4 Principle of Mathematical Induction Ex 4.1 Solutions
Question - 11 : - Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer - 11 : -
P (k + 1) is truewhenever P (k) is true.
Therefore, by theprinciple of mathematical induction, statement P (n) is true for all naturalnumbers i.e. n.
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer - 12 : -
P (k + 1) is truewhenever P (k) is true.
Therefore, by theprinciple of mathematical induction, statement P (n) is true for all naturalnumbers i.e. n.
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer - 13 : -
P (k + 1) is truewhenever P (k) is true.
Therefore, by theprinciple of mathematical induction, statement P (n) is true for all naturalnumbers i.e. n.
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer - 14 : -
By furthersimplification
= (k + 1) + 1
P (k + 1) is truewhenever P (k) is true.
Therefore, by theprinciple of mathematical induction, statement P (n) is true for all naturalnumbers i.e. n.
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer - 15 : -
P (k + 1) is true wheneverP (k) is true.
Therefore, by theprinciple of mathematical induction, statement P (n) is true for all naturalnumbers i.e. n.
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer - 16 : -
P (k + 1) is truewhenever P (k) is true.
Therefore, by theprinciple of mathematical induction, statement P (n) is true for all naturalnumbers i.e. n.
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer - 17 : -
P (k + 1) is truewhenever P (k) is true.
Therefore, by theprinciple of mathematical induction, statement P (n) is true for all naturalnumbers i.e. n.
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer - 18 : -
We can write the givenstatement as
P (k + 1) is truewhenever P (k) is true.
Therefore, by theprinciple of mathematical induction, statement P (n) is true for all naturalnumbers i.e. n.
Question - 19 : - Prove the following by using the principle of mathematical induction for all n ∈ N:
n (n + 1) (n + 5) is a multiple of 3
Answer - 19 : -
We can write the given statement as
P (n): n (n + 1) (n + 5), which is a multiple of 3
If n = 1 we get
1 (1 + 1) (1 + 5) = 12, which is a multiple of 3
Which is true.
Consider P (k) be true for some positive integer k
k (k + 1) (k + 5) is a multiple of 3
k (k + 1) (k + 5) = 3m, where m ∈ N …… (1)
Now let us prove that P (k + 1) is true.
Here
(k + 1) {(k + 1) + 1} {(k + 1) + 5}
We can write it as
= (k + 1) (k + 2) {(k + 5) + 1}
By multiplying the terms
= (k + 1) (k + 2) (k + 5) + (k + 1) (k + 2)
So we get
= {k (k + 1) (k + 5) + 2 (k + 1) (k + 5)} + (k + 1) (k + 2)
Substituting equation (1)
= 3m + (k + 1) {2 (k + 5) + (k + 2)}
By multiplication
= 3m + (k + 1) {2k + 10 + k + 2}
On further calculation
= 3m + (k + 1) (3k + 12)
Taking 3 as common
= 3m + 3 (k + 1) (k + 4)
We get
= 3 {m + (k + 1) (k + 4)}
= 3 × q where q = {m + (k + 1) (k + 4)} is some natural number
(k + 1) {(k + 1) + 1} {(k + 1) + 5} is a multiple of 3
P (k + 1) is true whenever P (k) is true.
Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.
Question - 20 : - Prove the following by using the principle of mathematical induction for all n ∈ N:
102n – 1 + 1 is divisible by 11
Answer - 20 : -
We can write the givenstatement as
P (n): 102n –1 + 1 is divisible by 11
If n = 1 we get
P (1) = 102.1 –1 + 1 = 11, which is divisible by 11
Which is true.
Consider P (k) be truefor some positive integer k
102k –1 + 1 is divisible by 11
102k –1 + 1 = 11m, where m ∈ N ……(1)
Now let us prove thatP (k + 1) is true.
Here
10 2 (k + 1)– 1 + 1
We can write it as
= 10 2k + 2– 1 + 1
= 10 2k + 1 +1
By addition andsubtraction of 1
= 10 2 (102k-1 +1 – 1) + 1
We get
= 10 2 (102k-1 +1) – 102 + 1
Using equation 1 weget
= 102. 11m– 100 + 1
= 100 × 11m – 99
Taking out the commonterms
= 11 (100m – 9)
= 11 r, where r =(100m – 9) is some natural number
10 2(k + 1)– 1 + 1 is divisible by 11
P (k + 1) is truewhenever P (k) is true.
Therefore, by theprinciple of mathematical induction, statement P (n) is true for all naturalnumbers i.e. n.