Chapter 4 Principle of Mathematical Induction Ex 4.1 Solutions
Question - 21 : - Prove the following by using the principle of mathematical induction for all n ∈ N:
x2n – y2n is divisible by x + y
Answer - 21 : -
We can write the givenstatement as
P (n): x2n – y2n isdivisible by x + y
If n = 1 we get
P (1) = x2 × 1 – y2 × 1 = x2 – y2 =(x + y) (x – y), which isdivisible by (x + y)
Which is true.
Consider P (k) be truefor some positive integer k
x2k – y2k is divisible by x + y
x2k – y2k = m (x + y),where m ∈ N …… (1)
Now let us prove thatP (k + 1) is true.
Here
x 2(k + 1) –y 2(k + 1)
We can write it as
= x 2k .x2 – y2k . y2
By adding andsubtracting y2k we get
= x2 (x2k –y2k + y2k) – y2k. y2
From equation (1) weget
= x2 {m(x + y) + y2k} – y2k. y2
By multiplying theterms
= m (x + y) x2 +y2k. x2 – y2k. y2
Taking out the commonterms
= m (x + y) x2 +y2k (x2 – y2)
Expanding usingformula
= m (x + y) x2 +y2k (x + y) (x – y)
So we get
= (x + y) {mx2 +y2k (x – y)}, which is a factor of (x + y)
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 - 22 : - Prove the following by using the principle of mathematical induction for all n ∈ N:
32n + 2 – 8n – 9 is divisible by8
Answer - 22 : -
We can write the givenstatement as
P (n): 32n +2 – 8n – 9 is divisible by 8
If n = 1 we get
P (1) = 32 × 1+ 2 – 8 × 1 – 9 = 64, which is divisible by 8
Which is true.
Consider P (k) be truefor some positive integer k
32k +2 – 8k – 9 is divisible by 8
32k +2 – 8k – 9 = 8m, where m ∈ N ……(1)
Now let us prove thatP (k + 1) is true.
Here
3 2(k + 1) +2 – 8 (k + 1) – 9
We can write it as
= 3 2k + 2 .32 – 8k – 8 – 9
By adding andsubtracting 8k and 9 we get
= 32 (32k+ 2 – 8k – 9 + 8k + 9) – 8k – 17
On furthersimplification
= 32 (32k+ 2 – 8k – 9) + 32 (8k + 9) – 8k – 17
From equation (1) weget
= 9. 8m + 9 (8k + 9) –8k – 17
By multiplying theterms
= 9. 8m + 72k + 81 –8k – 17
So we get
= 9. 8m + 64k + 64
By taking out thecommon terms
= 8 (9m + 8k + 8)
= 8r, where r = (9m +8k + 8) is a natural number
So 3 2(k +1) + 2 – 8 (k + 1) – 9 is divisible by 8
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 - 23 : - Prove the following by using the principle of mathematical induction for all n ∈ N:
41n – 14n is a multiple of27
Answer - 23 : -
We can write the givenstatement as
P (n):41n –14nis a multiple of 27
If n = 1 we get
P (1) = 411 –141 = 27, which is a multiple by 27
Which is true.
Consider P (k) be truefor some positive integer k
41k –14kis a multiple of 27
41k –14k = 27m, where m ∈ N ……(1)
Now let us prove thatP (k + 1) is true.
Here
41k + 1 –14 k + 1
We can write it as
= 41k. 41 –14k. 14
By adding andsubtracting 14k we get
= 41 (41k –14k + 14k) – 14k. 14
On furthersimplification
= 41 (41k –14k) + 41. 14k – 14k. 14
From equation (1) weget
= 41. 27m + 14k (41 – 14)
By multiplying theterms
= 41. 27m + 27. 14k
By taking out thecommon terms
= 27 (41m – 14k)
= 27r, where r = (41m– 14k) is a natural number
So 41k + 1 –14k + 1 is a multiple of 27
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 - 24 : - Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer - 24 : -
We can write the givenstatement as
P(n): (2n +7)< (n + 3)2
If n = 1 we get
2.1 + 7 = 9 < (1 +3)2 = 16
Which is true.
Consider P (k) be truefor some positive integer k
(2k + 7)< (k + 3)2 … (1)
Now let us prove thatP (k + 1) is true.
Here
{2 (k + 1) + 7} = (2k+ 7) + 2
We can write it as
= {2 (k + 1) + 7}
From equation (1) weget
(2k + 7) + 2 < (k +3)2 + 2
By expanding the terms
2 (k + 1) + 7 < k2 +6k + 9 + 2
On further calculation
2 (k + 1) + 7 < k2 +6k + 11
Here k2 +6k + 11 < k2 + 8k + 16
We can write it as
2 (k + 1) + 7 < (k+ 4)2
2 (k + 1) + 7 < {(k+ 1) + 3}2
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.