NCERT Solutions: Principle Of Mathematical Induction

# NCERT Solutions Class 11 Maths Chapter 4 - Principle Of Mathematical Induction

## NCERT QUESTION - (Principle Of Mathematical Induction)

Question 1:  Prove the following by using the principle of mathematical induction or all n ∈ N

ANSWER: Let the given statement be P(n), i.e.,

P(n): 1 +3+ 32  … 3n–1 =

For n = 1, we have

P(1): 1 =  , which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

1+ 3+ 32  … 3k–1  + 3(k 1) – 1

= (1 + 3+ 32  … 3k–1) + 3k

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 2: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

P(n):

For n = 1, we have

P(1): 13 = 1 =  , which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

13 + 23 + 33  …  +k3+  (k  + 1)3

= (13 + 23 + 33  ….  +k3)+ (k + 1)3

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 3: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

P(n):

For n = 1, we have

P(1): 1 =   which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 4:  Prove the following by using the principle of mathematical induction for all nN: 1.2.3 +2.3.4 …  n(n + 1) (n + 2) = ANSWER:  Let the given statement be P(n), i.e.,

P(n): 1.2.3+ 2.3.4 …  n(n + 1) (n+  2) =

For n = 1, we have

P(1): 1.2.3 = 6 =  , which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2.3 + 2.3.4 …  k(k + 1) (k  + 2)

We shall now prove that P(k + 1) is true.

Consider

1.2.3+ 2.3.4 …  +k(k + 1) (k + 2)+ (k + 1) (+k  2) (k + 3)

= {1.2.3 + 2.3.4 …  k(k + 1) (k + 2)} + (k + 1) (k + 2) (k + 3)

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 5:  Prove the following by using the principle of mathematical induction for all nN:

ANSWER:  Let the given statement be P(n), i.e.,

P(n) :

For n = 1, we have

P(1): 1.3 = 3  , which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

1.3+ 2.32 + 3.33  …  +k3k +(k  +1) 3k 1

= (1.3+ 2.32 + 3.33  …  +k.3k)+ (k + 1) 3k+ 1

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 6:Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

P(n):

For n = 1, we have

P(1):   , which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

1.2+ 2.3 +3.4 …  +k.(k+  1) +(k + 1).(k + 2)

= [1.2+ 2.3 +3.4 …  +k.(k  +1)] +(k + 1).(k + 2)

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 7: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

P(n):

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

(1.3+ 3.5+5.7 … (2k – 1) (2k + 1) +{2(k + 1) – 1}{2(k+  1) +1}

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 8: Prove the following by using the principle of mathematical induction for all nN: 1.2+ 2.22 + 3.22  …  +n.2n = (n – 1) 2n+ 1 + 2
ANSWER:  Let the given statement be P(n), i.e.,

P(n): 1.2+ 2.22 + 3.22+  …  +n.2n = (n – 1) 2n+ 1 + 2

For n = 1, we have

P(1): 1.2 = 2 = (1 – 1) 21+ 1+  2 = 0+ 2 = 2, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2 +2.22  +3.22 + … + k.2k = (k – 1) 2k+  1 + 2 … (i)

We shall now prove that P(k  +1) is true.

Consider

Thus, P(k  +1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 9: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

P(n):

For n = 1, we have

P(1):   , which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k  +1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 10: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

P(n):

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 11: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

P(n):

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 12: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 13: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

For n = 1, we have

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 14: Prove the following by using the principle of mathematical induction for all nN:

ANSWER:  Let the given statement be P(n), i.e.,

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 15: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 16:Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 17: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 18: Prove the following by using the principle of mathematical induction for all nN:
ANSWER:  Let the given statement be P(n), i.e.,

It can be noted that P(n) is true for n = 1 since   .

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Hence,

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 19: Prove the following by using the principle of mathematical induction for all nN: n (n+  1) (n + 5) is a multiple of 3.
ANSWER:  Let the given statement be P(n), i.e.,

P(n): n (n + 1) (n + 5), which is a multiple of 3.

It can be noted that P(n) is true for n = 1 since 1 (1+ 1) (1 +5) = 12, which is a multiple of 3.

Let P(k) be true for some positive integer k, i.e.,

k (k + 1) (k + 5) is a multiple of 3.

k (k + 1) (k  +5) = 3m, where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Thus, P(k  +1) is true whenever P(k) is true.

Hence, 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 nN: 102n – 1  +1 is divisible by 11.
ANSWER:  Let the given statement be P(n), i.e.,

P(n): 102n – 1  +1 is divisible by 11.

It can be observed that P(n) is true for n = 1 since P(1) = 102.1 – 1  +1 = 11, which is divisible by 11.

Let P(k) be true for some positive integer k, i.e.,

102k – 1 + 1 is divisible by 11.

∴102k – 1  +1 = 11m, where m ∈ … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 21: Prove the following by using the principle of mathematical induction for all nN: x2ny2n is divisible by x  + y.
ANSWER:  Let the given statement be P(n), i.e.,

P(n): x2ny2n is divisible by x  + y.

It can be observed that P(n) is true for n = 1.

This is so because x2 × 1 – y2 × 1x2 – y2 = ( + y) (x – y) is divisible by (xy).

Let P(k) be true for some positive integer k, i.e.,

x2k – y2k is divisible by x y.

x2k – y2km ( + y), where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 22: Prove the following by using the principle of mathematical induction for all nN: 32n+  2 – 8n – 9 is divisible by 8.
ANSWER:  Let the given statement be P(n), i.e.,

P(n): 32n+  2 – 8n – 9 is divisible by 8.

It can be observed that P(n) is true for n = 1 since 32 × 1 +2 – 8 × 1 – 9 = 64, which is divisible by 8.

Let P(k) be true for some positive integer k, i.e.,

32k+  2 – 8k – 9 is divisible by 8.

∴32k + 2 – 8k – 9 = 8m; where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 23: Prove the following by using the principle of mathematical induction for all nN: 41n – 14n is a multiple of 27.
ANSWER:  Let the given statement be P(n), i.e.,

P(n):41n – 14is a multiple of 27.

It can be observed that P(n) is true for n = 1 since  which is a multiple of 27.

Let P(k) be true for some positive integer k, i.e.,

41k – 14is a multiple of 27

∴41k – 14k = 27m, where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Thus, P(k  +1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question 24: Prove the following by using the principle of mathematical induction for all (2 +7) < (n + 3)2
ANSWER:  Let the given statement be P(n), i.e.,

P(n): (2n + 7) < (n + 3)2

It can be observed that P(n) is true for n = 1 since 2.1+ 7 = 9 < (1+ 3)2 = 16, which is true.

Let P(k) be true for some positive integer k, i.e.,

(2k + 7) < (k + 3)2 … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

The document NCERT Solutions Class 11 Maths Chapter 4 - Principle Of Mathematical Induction is a part of the Commerce Course Mathematics (Maths) Class 11.
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## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests

## FAQs on NCERT Solutions Class 11 Maths Chapter 4 - Principle Of Mathematical Induction

 1. How do you prove a statement using the principle of mathematical induction?
Ans. To prove a statement using the principle of mathematical induction, you first prove that the statement is true for the base case (usually n = 1). Then, you assume the statement is true for some arbitrary positive integer k, and use this assumption to prove that it must also be true for k+1. This completes the inductive step, and the statement is then proven to be true for all positive integers.
 2. What is the base case in a proof by mathematical induction?
Ans. The base case in a proof by mathematical induction is the initial step in the proof where you show that the statement holds true for the smallest possible value of the variable (usually n = 1). It is essential to establish the base case to start the induction process.
 3. Is mathematical induction the only way to prove mathematical statements?
Ans. No, mathematical induction is not the only way to prove mathematical statements. There are various other methods of proof, such as direct proof, proof by contradiction, proof by contrapositive, etc. Mathematical induction is specifically useful for proving statements that involve integers or natural numbers.
 4. Can mathematical induction be used to prove inequalities?
Ans. Yes, mathematical induction can be used to prove inequalities. When proving inequalities using mathematical induction, you would follow the same steps as for proving statements, but instead of showing that the statement holds true for a specific value, you would show that it holds true for all values greater than or equal to a certain value.
 5. How can mathematical induction be applied to solve problems in combinatorics or number theory?
Ans. Mathematical induction can be applied in combinatorics or number theory to prove various properties or relationships involving integers or counting principles. By formulating statements in terms of positive integers and proving them using mathematical induction, you can solve problems related to sequences, permutations, combinations, and other topics in these branches of mathematics.

## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests

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