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

**1 + 2 + 3 + … + n < ^{1}/_{8}(2n+1)^{2}**

Solution:

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

P(n): 1 + 2 + 3 + … + n < ^{1}/_{8}(2n+1)^{2}

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

1<^{1}/_{8 }(2.1 + 1)^{2} = ^{9}/_{8}

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

1 + 2 + 3 + … + k < ^{1}/_{8}(2k+1)^{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.

**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:

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.

**Prove the following by using the principle of mathematical induction for all n ∈ N: 10**^{2n – 1}+ 1 is divisible by 11.

Solution:

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

P(n): 10^{2n – 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., 10^{2k – 1} + 1 is divisible by 11.

∴ 10^{2k – 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 principal of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

**x ^{2n} – y^{2n} is divisible by x + y.**

Solution:

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

P(n): x^{2n} – y^{2n} is divisible by x + y.

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

This is so because x^{2 × 1} – y^{2 × 1} = x^{2} – y^{2} = (x + y) (x – y) is divisible by (x + y).

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

x^{2k} – y^{2k} is divisible by x + y.

∴ Let x^{2k} – y^{2k} = 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.

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

Solution:

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 3^{2 × 1 + 2} – 8 × 1 – 9 = 64, which is divisible by 8.

Let P(k) be true for some positive integer k, i.e., 3^{2k + 2} – 8k – 9 is divisible by 8.

∴ 3^{2k + 2} – 8k – 9 = 8m; where m ∈ N … (1) We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

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

** 41 ^{n} – 14^{n} is a multiple of 27.**

Solution:

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

P(n): 41^{n} – 14^{n} is a multiple of 27.

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

since 41^{l} – 14^{l} = 27, which is a multiple of 27.

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

41^{k} – 14^{k} is 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

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

**(2n +7) < (n + 3) ^{2}**

Solution:

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.