Principle of Mathematical Induction – Class XI – Exercise 4.1[Part 3]

18. 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.

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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.

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20. 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.

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

x²ⁿ – y²ⁿ is divisible by x + y.

Solution:

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 x^{2 × 1} – y^{2 × 1} = x² – y² = (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.

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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

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.

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

 41ⁿ – 14ⁿ is a multiple of 27.

Solution:

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

P(n): 41ⁿ – 14ⁿ 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

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24. 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.

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