Chapter 4. Principle of Mathematical induction
Exercise 4.1
Prove the following by using the principle of mathematical induction for all n ∈ N:
Solution:
Let the given statement be P(n), i.e.,
LHS = RHS
Thus P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.
Solution:
Let the given statement be P(n), i.e.,
LHS = RHS
Thus P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.
Solution: Let the given statement be P(n), so
Thus P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.
Q19. n (n + 1) (n + 5) is a multiple of 3.
Solution:
Let the given statement be P(n), so
P(n) : n (n + 1) (n + 5) is a multiple of 3.
For n = 1, so we have;
n (n + 1) (n + 5) = 1 × 2 × 6 = 12 = 3 × 4
P(n) is true for n = 1
Assume that P(k) is also true for some positive integer k.
k(k + 1) (k + 5)
= k3 + 6k2 + 5 k = 3m (say) ……………….. (1)
Now, we shall prove that P(k + 1) is true whenever P(k) is true
Replacing k by k + 1
k + 1 (k + 2) (k + 6)
= (k + 1) (k2 + 8k + 12)
= k (k2 + 8k + 12) + 1(k2 + 8k + 12)
= k3 + 8k2 + 12k + k2 + 8k + 12
= k3 + 9k2 + 20k + 12
=( k3 + 6k2 + 5 k) + 3k2 + 15k + 12
= 3m + 3k2 + 15k + 12 from (1)
= 3(m + k2 + 5k + 4)
∴ k + 1 (k + 2) (k + 6) is multiple of 3
Thus P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n ∈ N.
Q20. 102n - 1 + 1 is divisible by 11.
Solution:
Let the given statement be P(n), so
P(n) : 102n - 1 + 1 is divisible by 11.
For n = 1, so we have;
102n - 1 + 1 = 102×1 - 1 + 1 = 10 + 1 = 11
P(n) is true for n = 1
Assume that P(k) is also true for some positive integer k.
102k- 1 + 1 = 11m say
102k- 1 = 11m - 1 ……………… (1)
We shall prove that P(k + 1) is true whenever P(k) is true
∴ replacing k by k + 1 we have
102k - 1 + 1
= 102k + 1 + 1
= 102k × 101 + 1
= {102k - 1 × 100 + 1}
= {(11m - 1)× 100 + 1} from equation (1)
= 1100m - 100+ 1
= 1100m - 99
= 11(100m - 9)
∴ 102n - 1 + 1 is divisible by 11
Thus P(k + 1) is true, whenever P(k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n ∈ N.
Q21. x2n – y2n is divisible by x + y
Solution: Let the given statement be P(n), so
P(n) : x2n – y2n is divisible by x + y
Putting n = 1 we have,
x2n – y2n = x2 - y2 = (x + y) (x - y)
P(n) is true for n = 1
Assume that P(k) is also true for some positive integer k or
x2k – y2k is divisible by (x + y)
So, x2k – y2k = m( x + y)
Or x2k = m( x + y) + y2k …………. (1)
We shall prove that P(k + 1) is true whenever P(k) is true
∴ replacing k by k + 1 we have
x2k + 2 – y2k + 2
= x2k . x2 – y2k .y2
Putting the value of x2k from (1)
= {m( x + y) + y2k} x2 – y2k .y2
= m( x + y) x2 + y2k. x2 – y2k .y2
= m( x + y) x2 + y2k (x2 – y2)
= m( x + y) x2 + y2k (x + y) ( x - y)
= ( x + y) [mx2 + y2k ( x - y)]
∴ x2n – y2n is divisible by x + y
Thus P(k + 1) is true, whenever P(k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n ∈ N.
Q22. 32n+2 – 8n – 9 is divisible by 8
Solution: Let the given statement be P(n), so
P(n) : 32n+2 – 8n – 9 is divisible by 8
Putting n =1
P(1) : 32×1+2 – 8 × 1 – 9 = 81 - 17 = 64 = 8 × 8
Which is divisible by 8
∴ P(1) is true
Assume that P(k) is also true for some positive integer k
32k + 2 – 8k – 9
32k + 2 – 8k – 9 is divisible by 8
32k + 2 – 8k – 9 = 8m
Or 32k + 2 = 8m + 8k + 9 ……………. (1)
We shall prove that P(k + 1) is true whenever P(k) is true
∴ replacing k by k + 1 we have
32k + 4 – 8k – 8 – 9
= 32k + 4 – 8k – 17
= 32k + 2 × 32 – 8k – 17
= (8m + 8k + 9)× 9 – 8k – 17
= 72m + 72k + 81 – 8k – 17
= 72m + 64k + 64
= 8(9m + 8k + 8)
∴ 32n+2 – 8n – 9 is divisible by 8
Thus P(k + 1) is true, whenever P(k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n ∈ N.
Q23. 41n – 14n is a multiple of 27.
Solution: Let the given statement be P(n), so
P(n) : 41n – 14n is a multiple of 27
Putting n = 1
P(1): 41n – 14n = 41 – 14 = 27
∴ P(1) is true
Assume that P(k) is also true for some positive integer k
41k – 14k = 27
41k = 27 + 14k ………… (1)
We shall prove that P(k + 1) is true whenever P(k) is true
∴ replacing k by k + 1 we have
41k + 1 – 14k + 1
= 41k . 41 – 14k . 14
= (27 + 14k) 41 – 14k . 14
= 27 . 41 + 14k .41 – 14k . 14
= 27 . 41 + 14k (41 – 14 )
= 27 . 41 + 14k . 27
= 27 ( 41 + 14k )
∴ 41n – 14n is a multiple of 27
Thus P(k + 1) is true, whenever P(k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n ∈ N.
Q24. (2n + 7) < (n + 3)2
Solution: Let the statement be p(n) so,
p(n) : (2n + 7) < (n + 3)2
=> p(1) : (2 × 1 + 7) < (1 + 3)2
=> 9 < 42
=> 9 < 16
Therefore, p(1) is true so Assume that p(k) is also true for some integer k.
(2k + 7) < (k + 3)2 ......... (i)
Now we shall prove for p(k + 1)
2(k +1) + 7 < (k + 1 + 3)2
2k + 2 + 7 < (k + 4)2 ........ (ii)
We have from (i)
(2k + 7) < (k + 3)2
Adding 2 both sides
=> 2k + 7 + 2 < (k + 3)2 + 2
=> 2k + 7 + 2 < k2 + 6k + 9 + 2
=> 2k + 7 + 2 < k2 + 6k + 9 + 2
=> 2k + 7 + 2 < k2 + 6k + 11
Now, k2 + 6k + 11 < (k + 4)2 from (ii)
=> 2k + 7 + 2 < k2 + 6k + 11 < k2 + 8k + 16
=> 2k + 2 + 7 < k2 + 8k + 16
=> 2(k + 1) + 7 < (k + 4)2
=> 2(k + 1) + 7 < (k + 1 + 3)2
Thus P(k + 1) is true, whenever P(k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n ∈ N.
