RD Sharma Class 11 Solutions Chapter - Mathematical Induction | Mathematics (Maths) Class 11 - Commerce PDF Download

 Page 1


12. Mathematical Induction
Exercise 12.1
1. Question
If P(n) is the statement “n(n + 1) is even”, then what is P(3)?
Given. P(n) = n(n + 1) is even.
Find. P(3) ?
Answer
We have P(n) = n(n + 1).
= P(3) = 3(3 + 1)
= P(3) = 3(4)
Hence, P(3) = 12, So P(3) is also Even.
2. Question
If P(n) is the statement “n
3
 + n is divisible by 3”, prove that P(3) is true but P(4) is not true.
Answer
Given. P(n) = n
3
 + n is divisible by 3
Find P(3) is true but P(4) is not true
We have P(n) = n
3
 + n is divisible by 3
Let’s check with P(3)
= P(3) = 3
3
 + 3
= P(3) = 27 + 3
Therefore P(3) = 30, So it is divisible by 3
Now check with P(4)
= P(4) = 4
3
 + 4
= P(4) = 64 + 4
Therefore P(4) = 68, So it is not divisible by 3
Hence, P(3) is true and P(4) is not true.
3. Question
If P(n) is the statement “2
n
 = 3n”, and if P(r) is true, prove that P(r + 1) is true.
Answer
Given. P(n) = “2
n
 = 3n” and p(r) is true.
Prove. P(r + 1) is true
we have P(n) = 2
n
 = 3n
Since, P(r) is true So,
= 2
r
= 3r
Now, Multiply both side by 2
= 2.2
r
= 3r.2
Page 2


12. Mathematical Induction
Exercise 12.1
1. Question
If P(n) is the statement “n(n + 1) is even”, then what is P(3)?
Given. P(n) = n(n + 1) is even.
Find. P(3) ?
Answer
We have P(n) = n(n + 1).
= P(3) = 3(3 + 1)
= P(3) = 3(4)
Hence, P(3) = 12, So P(3) is also Even.
2. Question
If P(n) is the statement “n
3
 + n is divisible by 3”, prove that P(3) is true but P(4) is not true.
Answer
Given. P(n) = n
3
 + n is divisible by 3
Find P(3) is true but P(4) is not true
We have P(n) = n
3
 + n is divisible by 3
Let’s check with P(3)
= P(3) = 3
3
 + 3
= P(3) = 27 + 3
Therefore P(3) = 30, So it is divisible by 3
Now check with P(4)
= P(4) = 4
3
 + 4
= P(4) = 64 + 4
Therefore P(4) = 68, So it is not divisible by 3
Hence, P(3) is true and P(4) is not true.
3. Question
If P(n) is the statement “2
n
 = 3n”, and if P(r) is true, prove that P(r + 1) is true.
Answer
Given. P(n) = “2
n
 = 3n” and p(r) is true.
Prove. P(r + 1) is true
we have P(n) = 2
n
 = 3n
Since, P(r) is true So,
= 2
r
= 3r
Now, Multiply both side by 2
= 2.2
r
= 3r.2
= 2
r + 1
= 6r
= 2
r + 1
= 3r + 3r [since 3r>3 = 3r + 3r=3 + 3r]
Therefore 2
r + 1
= 3(r + 1)
Hence, P(r + 1) is true.
4. Question
If P(n) is the statement “n
2
 + n” is even”, and if P(r) is true, then P(r + 1) is true
Given. P(n) = n
2
 + n is even and P(r) is true.
Prove. P(r + 1) is true
Answer
Given P(r) is true that means,
= r
2
 + r is even
Let Assume r
2
 + r = 2k - - - - - - (i)
Now, (r + 1)
2
 + (r + 1)
r
2
 + 1 + 2r + r + 1
= (r
2
 + r) + 2r + 2
= 2k + 2r + 2
= 2(k + r + 1)
= 2µ
Therefore, (r + 1)
2
 + (r + 1) is Even.
Hence, P(r + 1) is true
5. Question
Given an example of a statement P(n) such that it is true for all n ? N.
Answer
P(n) = 1 + 2 + 3 + - - - - - + n = 
P(n) is true for all natural numbers.
Hence, P(n) is true for all n?N
6. Question
If P(n) is the statement “n
2
 – n + 41 is prime”, prove that P(1), P(2) and P(3) are true. Prove also that P(41) is
not true.
Given. P(n) = n
2
 - n + 41 is prime
Prove. P(1),P(2) and P(3) are true and P(41) is not true.
Answer
P(n) = n
2
 - n + 41
= P(1) = 1 - 1 + 41
= P(1) = 41
Therefore, P(1) is Prime
Page 3


12. Mathematical Induction
Exercise 12.1
1. Question
If P(n) is the statement “n(n + 1) is even”, then what is P(3)?
Given. P(n) = n(n + 1) is even.
Find. P(3) ?
Answer
We have P(n) = n(n + 1).
= P(3) = 3(3 + 1)
= P(3) = 3(4)
Hence, P(3) = 12, So P(3) is also Even.
2. Question
If P(n) is the statement “n
3
 + n is divisible by 3”, prove that P(3) is true but P(4) is not true.
Answer
Given. P(n) = n
3
 + n is divisible by 3
Find P(3) is true but P(4) is not true
We have P(n) = n
3
 + n is divisible by 3
Let’s check with P(3)
= P(3) = 3
3
 + 3
= P(3) = 27 + 3
Therefore P(3) = 30, So it is divisible by 3
Now check with P(4)
= P(4) = 4
3
 + 4
= P(4) = 64 + 4
Therefore P(4) = 68, So it is not divisible by 3
Hence, P(3) is true and P(4) is not true.
3. Question
If P(n) is the statement “2
n
 = 3n”, and if P(r) is true, prove that P(r + 1) is true.
Answer
Given. P(n) = “2
n
 = 3n” and p(r) is true.
Prove. P(r + 1) is true
we have P(n) = 2
n
 = 3n
Since, P(r) is true So,
= 2
r
= 3r
Now, Multiply both side by 2
= 2.2
r
= 3r.2
= 2
r + 1
= 6r
= 2
r + 1
= 3r + 3r [since 3r>3 = 3r + 3r=3 + 3r]
Therefore 2
r + 1
= 3(r + 1)
Hence, P(r + 1) is true.
4. Question
If P(n) is the statement “n
2
 + n” is even”, and if P(r) is true, then P(r + 1) is true
Given. P(n) = n
2
 + n is even and P(r) is true.
Prove. P(r + 1) is true
Answer
Given P(r) is true that means,
= r
2
 + r is even
Let Assume r
2
 + r = 2k - - - - - - (i)
Now, (r + 1)
2
 + (r + 1)
r
2
 + 1 + 2r + r + 1
= (r
2
 + r) + 2r + 2
= 2k + 2r + 2
= 2(k + r + 1)
= 2µ
Therefore, (r + 1)
2
 + (r + 1) is Even.
Hence, P(r + 1) is true
5. Question
Given an example of a statement P(n) such that it is true for all n ? N.
Answer
P(n) = 1 + 2 + 3 + - - - - - + n = 
P(n) is true for all natural numbers.
Hence, P(n) is true for all n?N
6. Question
If P(n) is the statement “n
2
 – n + 41 is prime”, prove that P(1), P(2) and P(3) are true. Prove also that P(41) is
not true.
Given. P(n) = n
2
 - n + 41 is prime
Prove. P(1),P(2) and P(3) are true and P(41) is not true.
Answer
P(n) = n
2
 - n + 41
= P(1) = 1 - 1 + 41
= P(1) = 41
Therefore, P(1) is Prime
= P(2) = 2
2
 – 2 + 41
= P(2) = 4 - 2 + 41
= P(2) = 43
Therefore, P(2) is prime
= P(3) = 3
2
 – 3 + 41
= P(3) = 9 – 3 + 41
= P(3) = 47
Therefore P(3) is prime
Now, P(41) = (41)
2
 - 41 + 41
= P(41) = 1681
Therefore, P(41) is not prime
Hence, P(1),P(2),P(3) are true but P(41) is not true.
Exercise 12.2
1. Question
Prove the following by the principle of mathematical induction:
 i.e., the sum of the first n natural numbers is 
Answer
Let us Assume P(n) = 1 + 2 + 3 + - - - - - - + n = 
For n = 1
L.H.S of P(n) = 1
R.H.S of P(n) =  = 1
Therefore, L.H.S =R.H.S
Since, P(n) is true for n = 1
Let assume P(n) be the true for n = k, so
1 + 2 + 3 + - - - - - + k =  - - - (1)
Now
(1 + 2 + 3 + - - + k) + (k + 1)
=  + (k + 1)
= (k + 1)
= 
= 
P(n) is true for n = k + 1
P(n) is true for all n?N
Page 4


12. Mathematical Induction
Exercise 12.1
1. Question
If P(n) is the statement “n(n + 1) is even”, then what is P(3)?
Given. P(n) = n(n + 1) is even.
Find. P(3) ?
Answer
We have P(n) = n(n + 1).
= P(3) = 3(3 + 1)
= P(3) = 3(4)
Hence, P(3) = 12, So P(3) is also Even.
2. Question
If P(n) is the statement “n
3
 + n is divisible by 3”, prove that P(3) is true but P(4) is not true.
Answer
Given. P(n) = n
3
 + n is divisible by 3
Find P(3) is true but P(4) is not true
We have P(n) = n
3
 + n is divisible by 3
Let’s check with P(3)
= P(3) = 3
3
 + 3
= P(3) = 27 + 3
Therefore P(3) = 30, So it is divisible by 3
Now check with P(4)
= P(4) = 4
3
 + 4
= P(4) = 64 + 4
Therefore P(4) = 68, So it is not divisible by 3
Hence, P(3) is true and P(4) is not true.
3. Question
If P(n) is the statement “2
n
 = 3n”, and if P(r) is true, prove that P(r + 1) is true.
Answer
Given. P(n) = “2
n
 = 3n” and p(r) is true.
Prove. P(r + 1) is true
we have P(n) = 2
n
 = 3n
Since, P(r) is true So,
= 2
r
= 3r
Now, Multiply both side by 2
= 2.2
r
= 3r.2
= 2
r + 1
= 6r
= 2
r + 1
= 3r + 3r [since 3r>3 = 3r + 3r=3 + 3r]
Therefore 2
r + 1
= 3(r + 1)
Hence, P(r + 1) is true.
4. Question
If P(n) is the statement “n
2
 + n” is even”, and if P(r) is true, then P(r + 1) is true
Given. P(n) = n
2
 + n is even and P(r) is true.
Prove. P(r + 1) is true
Answer
Given P(r) is true that means,
= r
2
 + r is even
Let Assume r
2
 + r = 2k - - - - - - (i)
Now, (r + 1)
2
 + (r + 1)
r
2
 + 1 + 2r + r + 1
= (r
2
 + r) + 2r + 2
= 2k + 2r + 2
= 2(k + r + 1)
= 2µ
Therefore, (r + 1)
2
 + (r + 1) is Even.
Hence, P(r + 1) is true
5. Question
Given an example of a statement P(n) such that it is true for all n ? N.
Answer
P(n) = 1 + 2 + 3 + - - - - - + n = 
P(n) is true for all natural numbers.
Hence, P(n) is true for all n?N
6. Question
If P(n) is the statement “n
2
 – n + 41 is prime”, prove that P(1), P(2) and P(3) are true. Prove also that P(41) is
not true.
Given. P(n) = n
2
 - n + 41 is prime
Prove. P(1),P(2) and P(3) are true and P(41) is not true.
Answer
P(n) = n
2
 - n + 41
= P(1) = 1 - 1 + 41
= P(1) = 41
Therefore, P(1) is Prime
= P(2) = 2
2
 – 2 + 41
= P(2) = 4 - 2 + 41
= P(2) = 43
Therefore, P(2) is prime
= P(3) = 3
2
 – 3 + 41
= P(3) = 9 – 3 + 41
= P(3) = 47
Therefore P(3) is prime
Now, P(41) = (41)
2
 - 41 + 41
= P(41) = 1681
Therefore, P(41) is not prime
Hence, P(1),P(2),P(3) are true but P(41) is not true.
Exercise 12.2
1. Question
Prove the following by the principle of mathematical induction:
 i.e., the sum of the first n natural numbers is 
Answer
Let us Assume P(n) = 1 + 2 + 3 + - - - - - - + n = 
For n = 1
L.H.S of P(n) = 1
R.H.S of P(n) =  = 1
Therefore, L.H.S =R.H.S
Since, P(n) is true for n = 1
Let assume P(n) be the true for n = k, so
1 + 2 + 3 + - - - - - + k =  - - - (1)
Now
(1 + 2 + 3 + - - + k) + (k + 1)
=  + (k + 1)
= (k + 1)
= 
= 
P(n) is true for n = k + 1
P(n) is true for all n?N
So , by the principle of Mathematical Induction
Hence, P(n) = 1 + 2 + 3 + - - - + n =  is true for all n?N
2. Question
Prove the following by the principle of mathematical induction:
To prove: Prove that by the Mathematical Induction.
Answer
Let Assume P(n):1
2
 + 2
2
 + 3
2
 + - - - + n
2
 = 
For n = 1
P(1): 1 = 
1=1
= P(n) is true for n = 1
Let P(n) is true for n = k, so
P(k): 1
2
 + 2
2
 + 3
2
 + - - - - - + k
2
 = 
Let’s check for P(n) = k + 1, So
P(k): 1
2
 + 2
2
 + 3
2
 + - - - - - + k
2
 + (k + 1)
2
 = 
= 1
2
 + 2
2
 + 3
2
 + - - - - - + k
2
 + (k + 1)
2
= 
= 
= 
= 
= 
= 
= 
Therefore, P(n) is true for n = k + 1
Hence, P(n) is true for all n?N by PMI
3. Question
Prove the following by the principle of mathematical induction:
Page 5


12. Mathematical Induction
Exercise 12.1
1. Question
If P(n) is the statement “n(n + 1) is even”, then what is P(3)?
Given. P(n) = n(n + 1) is even.
Find. P(3) ?
Answer
We have P(n) = n(n + 1).
= P(3) = 3(3 + 1)
= P(3) = 3(4)
Hence, P(3) = 12, So P(3) is also Even.
2. Question
If P(n) is the statement “n
3
 + n is divisible by 3”, prove that P(3) is true but P(4) is not true.
Answer
Given. P(n) = n
3
 + n is divisible by 3
Find P(3) is true but P(4) is not true
We have P(n) = n
3
 + n is divisible by 3
Let’s check with P(3)
= P(3) = 3
3
 + 3
= P(3) = 27 + 3
Therefore P(3) = 30, So it is divisible by 3
Now check with P(4)
= P(4) = 4
3
 + 4
= P(4) = 64 + 4
Therefore P(4) = 68, So it is not divisible by 3
Hence, P(3) is true and P(4) is not true.
3. Question
If P(n) is the statement “2
n
 = 3n”, and if P(r) is true, prove that P(r + 1) is true.
Answer
Given. P(n) = “2
n
 = 3n” and p(r) is true.
Prove. P(r + 1) is true
we have P(n) = 2
n
 = 3n
Since, P(r) is true So,
= 2
r
= 3r
Now, Multiply both side by 2
= 2.2
r
= 3r.2
= 2
r + 1
= 6r
= 2
r + 1
= 3r + 3r [since 3r>3 = 3r + 3r=3 + 3r]
Therefore 2
r + 1
= 3(r + 1)
Hence, P(r + 1) is true.
4. Question
If P(n) is the statement “n
2
 + n” is even”, and if P(r) is true, then P(r + 1) is true
Given. P(n) = n
2
 + n is even and P(r) is true.
Prove. P(r + 1) is true
Answer
Given P(r) is true that means,
= r
2
 + r is even
Let Assume r
2
 + r = 2k - - - - - - (i)
Now, (r + 1)
2
 + (r + 1)
r
2
 + 1 + 2r + r + 1
= (r
2
 + r) + 2r + 2
= 2k + 2r + 2
= 2(k + r + 1)
= 2µ
Therefore, (r + 1)
2
 + (r + 1) is Even.
Hence, P(r + 1) is true
5. Question
Given an example of a statement P(n) such that it is true for all n ? N.
Answer
P(n) = 1 + 2 + 3 + - - - - - + n = 
P(n) is true for all natural numbers.
Hence, P(n) is true for all n?N
6. Question
If P(n) is the statement “n
2
 – n + 41 is prime”, prove that P(1), P(2) and P(3) are true. Prove also that P(41) is
not true.
Given. P(n) = n
2
 - n + 41 is prime
Prove. P(1),P(2) and P(3) are true and P(41) is not true.
Answer
P(n) = n
2
 - n + 41
= P(1) = 1 - 1 + 41
= P(1) = 41
Therefore, P(1) is Prime
= P(2) = 2
2
 – 2 + 41
= P(2) = 4 - 2 + 41
= P(2) = 43
Therefore, P(2) is prime
= P(3) = 3
2
 – 3 + 41
= P(3) = 9 – 3 + 41
= P(3) = 47
Therefore P(3) is prime
Now, P(41) = (41)
2
 - 41 + 41
= P(41) = 1681
Therefore, P(41) is not prime
Hence, P(1),P(2),P(3) are true but P(41) is not true.
Exercise 12.2
1. Question
Prove the following by the principle of mathematical induction:
 i.e., the sum of the first n natural numbers is 
Answer
Let us Assume P(n) = 1 + 2 + 3 + - - - - - - + n = 
For n = 1
L.H.S of P(n) = 1
R.H.S of P(n) =  = 1
Therefore, L.H.S =R.H.S
Since, P(n) is true for n = 1
Let assume P(n) be the true for n = k, so
1 + 2 + 3 + - - - - - + k =  - - - (1)
Now
(1 + 2 + 3 + - - + k) + (k + 1)
=  + (k + 1)
= (k + 1)
= 
= 
P(n) is true for n = k + 1
P(n) is true for all n?N
So , by the principle of Mathematical Induction
Hence, P(n) = 1 + 2 + 3 + - - - + n =  is true for all n?N
2. Question
Prove the following by the principle of mathematical induction:
To prove: Prove that by the Mathematical Induction.
Answer
Let Assume P(n):1
2
 + 2
2
 + 3
2
 + - - - + n
2
 = 
For n = 1
P(1): 1 = 
1=1
= P(n) is true for n = 1
Let P(n) is true for n = k, so
P(k): 1
2
 + 2
2
 + 3
2
 + - - - - - + k
2
 = 
Let’s check for P(n) = k + 1, So
P(k): 1
2
 + 2
2
 + 3
2
 + - - - - - + k
2
 + (k + 1)
2
 = 
= 1
2
 + 2
2
 + 3
2
 + - - - - - + k
2
 + (k + 1)
2
= 
= 
= 
= 
= 
= 
= 
Therefore, P(n) is true for n = k + 1
Hence, P(n) is true for all n?N by PMI
3. Question
Prove the following by the principle of mathematical induction:
Answer
Let P(n) : 1 + 3 + 3
2
 + - - - - + 3
n - 1
 = 
Now, For n =1
P(1): 1 = =1
Therefore, P(n) is true for n =1
Now , P(n) is true for n = k
P(k) : 1 + 3 + 3
2
 + - - - - + 3
k - 1
 =  - - - - - (1)
Now, We have to show P(n) is true for n = k + 1
i.e P(k + 1): 1 + 3 + 3
2
 + - - - - + 3
k
 = 
then, {1 + 3 + 3
2
 + - - - - + 3
k - 1
} + 3
k + 1 - 1
=  using equation (1)
= 
= 
= 
Therefore, P(n) is true for n = k + 1
Hence, P(n) is true for all n?N
4. Question
Prove the following by the principle of mathematical induction:
Answer
Let P(n): 
For n = 1
P(1): 
= P(n) is true for n = 1
Let P(n) is true for n = k, So
 - - - - - (1)
Now, Let P(n) is true for n = k + 1, So