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Question 1:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 1:

Let the given statement be p(n)

p(n):

for n=1

p(1):1=(3-1)/2=2/2=1,which is true

Assume that p(k)is true for some positive integer k. ie

1+3+3²+....+ $3^{k-1}$ =( $3^{k}$ -1)/2 .......................(1)

p(k+1): 1+3+3²+....+ $3^{k-1}$ + $3^{k}$ =( $3^{k}$ -1)/2+ $3^{k}$ (using (1))

=( $3^{k}$ -1+2* $3^{k}$ )/2

=(3* $3^{k}$ -1)/2

=( $3^{k+1}$ -1)/2

p(k+1) is true,if p(k) is true.

therefore by the principle of mathematical induction,given statement p(n) is true for all natural numbers.

Comments

Afeefa T.

Question 2:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 2:

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.

Comments

Swapna Shree

Question 3:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 3:

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.

Comments

Swapna Shree

Question 4:

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =

Solution 4:

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.

Comments

Swapna Shree

Question 5:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 5:

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 + … + k3^k+ (k + 1) 3^k⁺¹

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

Comments

Swapna Shree

Question 6:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 6:

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.

Comments

Swapna Shree

Question 7:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 7:

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.

Comments

Swapna Shree

Question 8:

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2

Solution 8:

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.2^k = (k – 1) 2^k^{+ 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.

Comments

Swapna Shree

Question 9:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 9:

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.

Comments

Swapna Shree

Question 10:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 10:

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.

Comments

Swapna Shree

Question 11:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 11:

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.

Comments

Swapna Shree

Question 12:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 12:

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.

Comments

Swapna Shree

Question 13:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 13:

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.

Comments

Swapna Shree

Question 14:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 14:

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.

Comments

Swapna Shree

Question 15:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 15:

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.

Comments

Swapna Shree

Question 16:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 16:

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.

Comments

Swapna Shree

Question 17:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 17:

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.

Comments

Swapna Shree

Question 18:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution 18:

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.

Comments

Swapna Shree

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.

Solution 19:

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.

Comments

Swapna Shree

Question 20:

Prove the following by using the principle of mathematical induction for all n ∈ N: 10²ⁿ^{– 1}+ 1 is divisible by 11.

Solution 20:

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) = 10^{2.1 – 1}+ 1 = 11, which is divisible by 11.

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

10²^k^{– 1}+ 1 is divisible by 11.

∴10²^k^{– 1}+ 1 = 11m, 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.

Comments

Swapna Shree

Question 21:

Prove the following by using the principle of mathematical induction for all n ∈ N: x²ⁿ – y²ⁿ is divisible by x + y.

Solution 21:

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

P(n): x²ⁿ – y²ⁿ is divisible by x + y.

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

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

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

x²^k – y²^k is divisible by x + y.

∴x²^k – y²^k = m (x + 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.

Comments

Swapna Shree

Question 22:

Prove the following by using the principle of mathematical induction for all n ∈ N: 3²ⁿ^{+ 2} – 8n – 9 is divisible by 8.

Solution 22:

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.,

3²^k^{+ 2} – 8k – 9 is divisible by 8.

∴3²^k^{+ 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.

Comments

Swapna Shree

Question 23:

Prove the following by using the principle of mathematical induction for all n ∈ N: 41ⁿ – 14ⁿ is a multiple of 27.

Solution 23:

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

P(n):41ⁿ – 14ⁿis 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.,

41^k – 14^kis a multiple of 27

∴41^k – 14^k = 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.

Comments

Swapna Shree

Question 24:

Prove the following by using the principle of mathematical induction for all

(2n +7) < (n + 3)²

Solution 24:

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.

Comments

Swapna Shree

All Exercises - Chapter 4 - Principle of Mathematical Induction

Exercise 4.1

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All Exercises - Chapter 4 - Principle of Mathematical Induction

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