RD Sharma Solutions for Class 11 Maths Chapter 12 Mathematical Induction

RD Sharma Solutions Class 11 Maths Chapter 12 – Free PDF Download

RD Sharma Solutions for Class 11 Maths Chapter 12 – Mathematical Induction are provided here for students to score good marks in the board exams. Problems pertaining to principles of mathematical induction are dealt with in this chapter. The solutions to this chapter are formulated by our subject expert experts to boost students’ confidence levels in understanding the concepts and methods to solve problems in a shorter period. RD Sharma Class 11 Maths Solutions help students who aim to secure a good academic score in the board exams. 

Chapter 12 – Mathematical Induction contains two exercises. This chapter of RD Sharma Class 11 mainly focuses on the concept of mathematical induction and provides precise answers to each exercise. To know more about this topic, students can download the  RD Sharma Solutions for Class 11 Maths Chapter 12 updated for the 2023-24 exam from the links given below. Now, let us have a look at the concepts discussed in this chapter.

• Mathematical statements
• The principles of mathematical induction
  • The first principle of mathematical induction
  • The second principle of mathematical induction

RD Sharma Solutions for Class 11 Maths Chapter 12 – Mathematical Induction

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Exercise 12.1 Solutions

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EXERCISE 12.1 PAGE NO: 12.3

1. If P (n) is the statement “n (n + 1) is even”, then what is P (3)?
Solution:

Given:

P (n) = n (n + 1) is even.

So,

P (3) = 3 (3 + 1)

= 3 (4)

= 12

Hence, P (3) = 12, P (3) is also even.

2. If P (n) is the statement “n³ + n is divisible by 3”, prove that P (3) is true, but P (4) is not true.

Solution:

Given:

P (n) = n³ + n is divisible by 3.

We have P (n) = n³ + n

So,

P (3) = 3³ + 3

= 27 + 3

= 30

P (3) = 30, so it is divisible by 3.

Now, let’s check with P (4).

P (4) = 4³ + 4

= 64 + 4

= 68

P (4) = 68, so it is not divisible by 3

Hence, P (3) is true, and P (4) is not true.

3. If P (n) is the statement “2ⁿ ≥ 3n”, and if P (r) is true, prove that P (r + 1) is true.

Solution:

Given:

P (n) = “2ⁿ ≥ 3n” and p(r) is true.

We have P (n) = 2ⁿ ≥ 3n

Since P (r) is true

So,

2^r≥ 3r

Now, let’s multiply both sides by 2.

2×2^r≥ 3r×2

2^{r + 1}≥ 6r

2^{r + 1}≥ 3r + 3r [since 3r>3 = 3r + 3r≥3 + 3r]

∴ 2^{r + 1}≥ 3(r + 1)

Hence, P (r + 1) is true.

4. If P (n) is the statement “n² + n” is even”, and if P (r) is true, then P (r + 1) is true
Solution:

Given:

P (n) = n² + n is even, and P (r) is true, then r² + r is even.

Let us consider r² + r = 2k … (i)

Now, (r + 1)² + (r + 1)

r² + 1 + 2r + r + 1

(r² + r) + 2r + 2

2k + 2r + 2 [from equation (i)]

2(k + r + 1)

2μ

∴ (r + 1)² + (r + 1) is Even.

Hence, P (r + 1) is true.

5. Given an example of a statement P (n) such that it is true for all n ϵ N.

Solution:

Let us consider

P (n) = 1 + 2 + 3 + – – – – – + n = n(n+1)/2

So,

P (n) is true for all natural numbers.

Hence, P (n) is true for all n ∈ N.

6. If P (n) is the statement “n² – n + 41 is prime”, prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.
Solution:

Given:

P(n) = n² – n + 41 is prime.

P(n) = n² – n + 41

P (1) = 1 – 1 + 41

= 41

P (1) is Prime.

Similarly,

P(2) = 2² – 2 + 41

= 4 – 2 + 41

= 43

P (2) is prime.

Similarly,

P (3) = 3² – 3 + 41

= 9 – 3 + 41

= 47

P (3) is prime

Now,

P (41) = (41)² – 41 + 41

= 1681

P (41) is not prime

Hence, P (1), P(2), P (3) are true, but P (41) is not true.

EXERCISE 12.2 PAGE NO: 12.27

Prove the following by the principle of mathematical induction:

1. 1 + 2 + 3 + … + n = n (n +1)/2, i.e., the sum of the first n natural numbers is n (n + 1)/2.

Solution:

Let us consider P (n) = 1 + 2 + 3 + ….. + n = n (n +1)/2

For, n = 1

LHS of P (n) = 1

RHS of P (n) = 1 (1+1)/2 = 1

So, LHS = RHS

Since P (n) is true for n = 1

Let us consider P (n) to be true for n = k, so

1 + 2 + 3 + …. + k = k (k+1)/2 … (i)

Now,

(1 + 2 + 3 + … + k) + (k + 1) = k (k+1)/2 + (k+1)

= (k + 1) (k/2 + 1)

= [(k + 1) (k + 2)] / 2

= [(k+1) [(k+1) + 1]] / 2

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 = n (n +1)/2 is true for all n ∈ N.

2. 1² + 2² + 3² + … + n² = [n (n+1) (2n+1)]/6

Solution:

Let us consider P (n) = 1² + 2² + 3² + … + n² = [n (n+1) (2n+1)]/6

For, n = 1

P (1) = [1 (1+1) (2+1)]/6

1 = 1

P (n) is true for n = 1

Let P (n) is true for n = k, so

P (k): 1² + 2² + 3² + … + k² = [k (k+1) (2k+1)]/6

Let’s check for P (n) = k + 1, so

P (k) = 1² + 2² + 3² + – – – – – + k² + (k + 1)² = [k + 1 (k+2) (2k+3)] /6

= 1² + 2² + 3² + – – – – – + k² + (k + 1)²

= [k + 1 (k+2) (2k+3)] /6 + (k + 1)²

= (k +1) [(2k² + k)/6 + (k + 1)/1]

= (k +1) [2k² + k + 6k + 6]/6

= (k +1) [2k² + 7k + 6]/6

= (k +1) [2k² + 4k + 3k + 6]/6

= (k +1) [2k(k + 2) + 3(k + 2)]/6

= [(k +1) (2k + 3) (k + 2)] / 6

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

3. 1 + 3 + 3² + … + 3^n-1 = (3ⁿ – 1)/2

Solution:

Let P (n) = 1 + 3 + 3² + – – – – + 3^{n – 1} = (3ⁿ – 1)/2

Now, For n = 1

P (1) = 1 = (3¹ – 1)/2 = 2/2 =1

P (n) is true for n = 1

Now, let’s check that P (n) is true for n = k.

P (k) = 1 + 3 + 3² + – – – – + 3^{k – 1} = (3^k – 1)/2 … (i)

Now, we have to show that P (n) is true for n = k + 1

P (k + 1) = 1 + 3 + 3² + – – – – + 3^k = (3^k+1 – 1)/2

Then, {1 + 3 + 3² + – – – – + 3^{k – 1}} + 3^{k + 1 – 1}

= (3k – 1)/2 + 3^k using equation (i),

= (3k – 1 + 2×3^k)/2

= (3×3 k – 1)/2

= (3^k+1 – 1)/2

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

4. 1/1.2 + 1/2.3 + 1/3.4 + … + 1/n(n+1) = n/(n+1)

Solution:

Let P (n) = 1/1.2 + 1/2.3 + 1/3.4 + … + 1/n(n+1) = n/(n+1)

For, n = 1

P (n) = 1/1.2 = 1/1+1

1/2 = 1/2

P (n) is true for n = 1

Let’s check for P (n) is true for n = k,

1/1.2 + 1/2.3 + 1/3.4 + … + 1/k(k+1) + k/(k+1) (k+2) = (k+1)/(k+2)

Then,

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

= 1/(k+1)/(k+2) + k/(k+1)

= 1/(k+1) [k(k+2)+1]/(k+2)

= 1/(k+1) [k² + 2k + 1]/(k+2)

=1/(k+1) [(k+1) (k+1)]/(k+2)

= (k+1) / (k+2)

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

5. 1 + 3 + 5 + … + (2n – 1) = n², i.e., the sum of the first n odd natural numbers is n².

Solution:

Let P (n): 1 + 3 + 5 + … + (2n – 1) = n²

Let us check if P (n) is true for n = 1.

P (1) = 1 =1²

1 = 1

P (n) is true for n = 1

Now, Let’s check that P (n) is true for n = k.

P (k) = 1 + 3 + 5 + … + (2k – 1) = k² … (i)

We have to show that

1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)²

Now,

1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1

= k² + (2k + 1)

= k² + 2k + 1

= (k + 1)²

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

6. 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3n-1) (3n+2) = n/(6n+4)

Solution:

Let P (n) = 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3n-1) (3n+2) = n/(6n+4)

Let us check if P (n) is true for n = 1

P (1): 1/2.5 = 1/6.1+4 => 1/10 = 1/10

P (1) is true.

Now,

Let us check that P (k) is true and prove that P (k + 1) is true.

P (k): 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3k-1) (3k+2) = k/(6k+4)

P (k +1): 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3k-1)(3k+2) + 1/(3k+3-1)(3k+3+2)

: k/(6k+4) + 1/(3k+2)(3k+5)

: [k(3k+5)+2] / [2(3k+2)(3k+5)]

: (k+1) / (6(k+1)+4)

P (k + 1) is true.

Hence proved by mathematical induction.

7. 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3n-2)(3n+1) = n/3n+1

Solution:

Let P (n) = 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3n-2)(3n+1) = n/3n+1

Let us check for n = 1,

P (1): 1/1.4 = 1/4

1/4 = 1/4

P (n) is true for n = 1.

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k) = 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3k-2)(3k+1) = k/3k+1 … (i)

So,

= k/(3k+1) + 1/(3k+1)(3k+4)

= 1/(3k+1) [k/1 + 1/(3k+4)]

= 1/(3k+1) [k(3k+4)+1]/(3k+4)

= 1/(3k+1) [3k² + 4k + 1]/ (3k+4)

= 1/(3k+1) [3k² + 3k+k+1]/(3k+4)

= [3k(k+1) + (k+1)] / [(3k+4) (3k+1)]

= [(3k+1)(k+1)] / [(3k+4) (3k+1)]

= (k+1) / (3k+4)

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

8. 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2n+1)(2n+3) = n/3(2n+3)

Solution:

Let P (n) = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2n+1)(2n+3) = n/3(2n+3)

Let us check for n = 1,

P (1): 1/3.5 = 1/3(2.1+3)

: 1/15 = 1/15

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k) = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) = k/3(2k+3) … (i)

So,

1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) + 1/[2(k+1)+1][2(k+1)+3]

1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) + 1/(2k+3)(2k+5)

Now, substituting the value of P (k), we get,

= k/3(2k+3) + 1/(2k+3)(2k+5)

= [k(2k+5)+3] / [3(2k+3)(2k+5)]

= (k+1) / [3(2(k+1)+3)]

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

9. 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4n-1)(4n+3) = n/3(4n+3)

Solution:

Let P (n) = 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4n-1)(4n+3) = n/3(4n+3)

Let us check for n = 1,

P (1): 1/3.7 = 1/(4.1-1)(4+3)

: 1/21 = 1/21

P (n) is true for n =1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k-1)(4k+3) = k/3(4k+3) …. (i)

So,

1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k-1)(4k+3) + 1/(4k+3)(4k+7)

Substituting the value of P (k), we get

= k/(4k+3) + 1/(4k+3)(4k+7)

= 1/(4k+3) [k(4k+7)+3] / [3(4k+7)]

= 1/(4k+3) [4k² + 7k +3]/ [3(4k+7)]

= 1/(4k+3) [4k² + 3k+4k+3] / [3(4k+7)]

= 1/(4k+3) [4k(k+1)+3(k+1)]/ [3(4k+7)]

= 1/(4k+3) [(4k+3)(k+1)] / [3(4k+7)]

= (k+1) / [3(4k+7)]

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

10. 1.2 + 2.2² + 3.2³ + … + n.2ⁿ = (n–1) 2^{n + 1} + 2

Solution:

Let P (n) = 1.2 + 2.2² + 3.2³ + … + n.2ⁿ = (n–1) 2^{n + 1} + 2

Let us check for n = 1,

P (1):1.2 = 0.2⁰ + 2

: 2 = 2

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1.2 + 2.2² + 3.2³ + … + k.2^k = (k–1) 2^{k + 1} + 2 …. (i)

So,

{1.2 + 2.2² + 3.2³ + … + k.2^k} + (k + 1)2^{k + 1}

Now, substituting the value of P (k), we get,

= [(k – 1)2^{k + 1} + 2] + (k + 1)2^{k + 1} using equation (i)

= (k – 1)2^{k + 1} + 2 + (k + 1)2^{k + 1}

= 2^{k + 1}(k – 1 + k + 1) + 2

= 2^{k + 1} × 2k + 2

= k × 2^{k + 2} + 2

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

11. 2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n (3n + 1)

Solution:

Let P (n) = 2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n (3n + 1)

Let us check for n = 1,

P (1): 2 = 1/2 × 1 × 4

: 2 = 2

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k) = 2 + 5 + 8 + 11 + … + (3k – 1) = 1/2 k (3k + 1) … (i)

So,

2 + 5 + 8 + 11 + … + (3k – 1) + (3k + 2)

Now, substituting the value of P (k), we get,

= 1/2 × k (3k + 1) + (3k + 2) by using equation (i)

= [3k² + k + 2 (3k + 2)] / 2

= [3k² + k + 6k + 2] / 2

= [3k² + 7k + 2] / 2

= [3k² + 4k + 3k + 2] / 2

= [3k (k+1) + 4(k+1)] / 2

= [(k+1) (3k+4)] /2

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

12. 1.3 + 2.4 + 3.5 + … + n. (n+2) = 1/6 n (n+1) (2n+7)

Solution:

Let P (n): 1.3 + 2.4 + 3.5 + … + n. (n+2) = 1/6 n (n+1) (2n+7)

Let us check for n = 1,

P (1): 1.3 = 1/6 × 1 × 2 × 9

: 3 = 3

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1.3 + 2.4 + 3.5 + … + k. (k+2) = 1/6 k (k+1) (2k+7) … (i)

So,

1.3 + 2.4 + 3.5 + … + k. (k+2) + (k+1) (k+3)

Now, substituting the value of P (k), we get,

= 1/6 k (k+1) (2k+7) + (k+1) (k+3) by using equation (i)

= (k+1) [{k(2k+7)/6} + {(k+3)/1}]

= (k+1) [(2k² + 7k + 6k + 18)] / 6

= (k+1) [2k² + 13k + 18] / 6

= (k+1) [2k² + 9k + 4k + 18] / 6

= (k+1) [2k(k+2) + 9(k+2)] / 6

= (k+1) [(2k+9) (k+2)] / 6

= 1/6 (k+1) (k+2) (2k+9)

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

13. 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = n(4n² + 6n – 1)/3

Solution:

Let P (n): 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = n(4n² + 6n – 1)/3

Let us check for n = 1,

P (1): (2.1 – 1) (2.1 + 1) = 1(4.1² + 6.1 -1)/3

: 1×3 = 1(4+6-1)/3

: 3 = 3

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) = k(4k² + 6k – 1)/3 … (i)

So,

1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1) (2k + 3)

Now, substituting the value of P (k), we get,

= k(4k² + 6k – 1)/3 + (2k + 1) (2k + 3) by using equation (i)

= [k(4k² + 6k-1) + 3 (4k² + 6k + 2k + 3)] / 3

= [4k³ + 6k² – k + 12k² + 18k + 6k + 9] /3

= [4k³ + 18k² + 23k + 9] /3

= [4k³ + 4k² + 14k² + 14k +9k + 9] /3

= [(k+1) (4k² + 8k +4 + 6k + 6 – 1)] / 3

= [(k+1) 4[(k+1)² + 6(k+1) -1]] /3

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

14. 1.2 + 2.3 + 3.4 + … + n(n+1) = [n (n+1) (n+2)] / 3

Solution:

Let P (n): 1.2 + 2.3 + 3.4 + … + n(n+1) = [n (n+1) (n+2)] / 3

Let us check for n = 1,

P (1): 1(1+1) = [1(1+1) (1+2)] /3

: 2 = 2

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1.2 + 2.3 + 3.4 + … + k(k+1) = [k (k+1) (k+2)] / 3 … (i)

So,

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

Now, substituting the value of P (k), we get,

= [k (k+1) (k+2)] / 3 + (k+1) (k+2) by using equation (i)

= (k+2) (k+1) [k/2 + 1]

= [(k+1) (k+2) (k+3)] /3

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

15. 1/2 + 1/4 + 1/8 + … + 1/2ⁿ = 1 – 1/2ⁿ

Solution:

Let P (n): 1/2 + 1/4 + 1/8 + … + 1/2ⁿ = 1 – 1/2ⁿ

Let us check for n = 1,

P (1): 1/2¹ = 1 – 1/2¹

: 1/2 = 1/2

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

Let P (k): 1/2 + 1/4 + 1/8 + … + 1/2^k = 1 – 1/2^k … (i)

So,

1/2 + 1/4 + 1/8 + … + 1/2^k + 1/2^k+1

Now, substituting the value of P (k), we get,

= 1 – 1/2^k + 1/2^k+1 by using equation (i)

= 1 – ((2-1)/2^k+1)

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

16. 1² + 3² + 5² + … + (2n – 1)² = 1/3 n (4n² – 1)

Solution:

Let P (n): 1² + 3² + 5² + … + (2n – 1)² = 1/3 n (4n² – 1)

Let us check for n = 1,

P (1): (2.1 – 1)² = 1/3 × 1 × (4 – 1)

: 1 = 1

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1² + 3² + 5² + … + (2k – 1)² = 1/3 k (4k² – 1) … (i)

So,

1² + 3² + 5² + … + (2k – 1)² + (2k + 1)²

Now, substituting the value of P (k), we get,

= 1/3 k (4k² – 1) + (2k + 1)² by using equation (i)

= 1/3 k (2k + 1) (2k – 1) + (2k + 1)²

= (2k + 1) [{k(2k-1)/3} + (2k+1)]

= (2k + 1) [2k² – k + 3(2k+1)] / 3

= (2k + 1) [2k² – k + 6k + 3] / 3

= [(2k+1) 2k² + 5k + 3] /3

= [(2k+1) (2k(k+1)) + 3 (k+1)] /3

= [(2k+1) (2k+3) (k+1)] /3

= (k+1)/3 [4k² + 6k + 2k + 3]

= (k+1)/3 [4k² + 8k – 1]

= (k+1)/3 [4(k+1)² – 1]

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

17. a + ar + ar² + … + ar^{n – 1} = a [(rⁿ – 1)/(r – 1)], r ≠ 1

Solution:

Let P (n): a + ar + ar² + … + ar^{n – 1} = a [(rⁿ – 1)/(r – 1)]

Let us check for n = 1,

P (1): a = a (r¹ – 1)/(r-1)

: a = a

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): a + ar + ar² + … + ar^{k – 1} = a [(r^k – 1)/(r – 1)] … (i)

So,

a + ar + ar² + … + ar^{k – 1} + ar^k

Now, substituting the value of P (k), we get,

= a [(r^k – 1)/(r – 1)] + ar^k by using equation (i)

= a[r^k – 1 + r^k(r-1)] / (r-1)

= a[r^k – 1 + r^k+1 – r^‑k] / (r-1)

= a[r^k+1 – 1] / (r-1)

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

18. a + (a + d) + (a + 2d) + … + (a + (n-1)d) = n/2 [2a + (n-1)d]

Solution:

Let P (n): a + (a + d) + (a + 2d) + … + (a + (n-1)d) = n/2 [2a + (n-1)d]

Let us check for n = 1,

P (1): a = ½ [2a + (1-1)d]

: a = a

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): a + (a + d) + (a + 2d) + … + (a + (k-1)d) = k/2 [2a + (k-1)d] … (i)

So,

a + (a + d) + (a + 2d) + … + (a + (k-1)d) + (a + (k)d)

Now, substituting the value of P (k), we get,

= k/2 [2a + (k-1)d] + (a + kd) by using equation (i)

= [2ka + k(k-1)d + 2(a+kd)] / 2

= [2ka + k²d – kd + 2a + 2kd] / 2

= [2ka + 2a + k²d + kd] / 2

= [2a(k+1) + d(k² + k)] / 2

= (k+1)/2 [2a + kd]

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

19. 5²ⁿ – 1 is divisible by 24 for all n ϵ N

Solution:

Let P (n): 5²ⁿ – 1 is divisible by 24

Let us check for n = 1,

P (1): 5² – 1 = 25 – 1 = 24

P (n) is true for n = 1. Where, P (n) is divisible by 24

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 5^2k – 1 is divisible by 24

: 5^2k – 1 = 24λ … (i)

We have to prove,

5^{2k + 1} – 1 is divisible by 24

5^{2(k + 1)} – 1 = 24μ

So,

= 5^{2(k + 1)} – 1

= 5^2k.5² – 1

= 25.5^2k – 1

= 25.(24λ + 1) – 1 by using equation (1)

= 25.24λ + 24

= 24λ

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

20. 3²ⁿ + 7 is divisible by 8 for all n ϵ N

Solution:

Let P (n): 3²ⁿ + 7 is divisible by 8

Let us check for n = 1,

P (1): 3² + 7 = 9 + 7 = 16

P (n) is true for n = 1. Where, P (n) is divisible by 8

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 3^2k + 7 is divisible by 8

: 3^2k + 7 = 8λ

: 3^2k = 8λ – 7 … (i)

We have to prove,

3^{2(k + 1)} + 7 is divisible by 8

3^{2k + 2} + 7 = 8μ

So,

= 3^{2(k + 1)} + 7

= 3^2k.3² + 7

= 9.3^2k + 7

= 9.(8λ – 7) + 7 by using equation (i)

= 72λ – 63 + 7

= 72λ – 56

= 8(9λ – 7)

= 8μ

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

21. 5^{2n + 2} – 24n – 25 is divisible by 576 for all n ϵ N

Solution:

Let P (n): 5^{2n + 2} – 24n – 25 is divisible by 576

Let us check for n = 1,

P (1): 5^2.1+2 – 24.1 – 25

: 625 – 49

: 576

P (n) is true for n = 1. Where, P (n) is divisible by 576

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 5^{2k + 2} – 24k – 25 is divisible by 576

: 5^{2k + 2} – 24k – 25 = 576λ …. (i)

We have to prove,

5^{2k + 4} – 24(k + 1) – 25 is divisible by 576

5^{(2k + 2) + 2} – 24(k + 1) – 25 = 576μ

So,

= 5^{(2k + 2) + 2} – 24(k + 1) – 25

= 5^{(2k + 2)}.5² – 24k – 24– 25

= (576λ + 24k + 25)25 – 24k– 49 by using equation (i),

= 25. 576λ + 576k + 576

= 576(25λ + k + 1)

= 576μ

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

22. 3^{2n + 2} – 8n – 9 is divisible by 8 for all n ϵ N

Solution:

Let P (n): 3^{2n + 2} – 8n – 9 is divisible by 8

Let us check for n = 1,

P (1): 3^{2.1 + 2} – 8.1 – 9

: 81 – 17

: 64

P (n) is true for n = 1. Where, P (n) is divisible by 8

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 3^{2k + 2} – 8k – 9 is divisible by 8.

: 3^{2k + 2} – 8k – 9 = 8λ … (i)

We have to prove,

3^{2k + 4} – 8(k + 1) – 9 is divisible by 8.

3^{(2k + 2) + 2} – 8(k + 1) – 9 = 8μ

So,

= 3^{2(k + 1)}.3² – 8(k + 1) – 9

= (8λ + 8k + 9)9 – 8k – 8 – 9

= 72λ + 72k + 81 – 8k – 17 using equation (1),

= 72λ + 64k + 64

= 8(9λ + 8k + 8)

= 8μ

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

23. (ab) ⁿ = aⁿ bⁿ for all n ϵ N

Solution:

Let P (n): (ab) ⁿ = aⁿ bⁿ

Let us check for n = 1,

P (1): (ab) ¹ = a¹ b¹

: ab = ab

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): (ab) ^k = a^k b^k … (i)

We have to prove,

(ab) ^{k + 1} = a^{k + 1}.b^{k + 1}

So,

= (ab) ^{k + 1}

= (ab) ^k (ab)

= (a^k b^k) (ab) using equation (1)

= (a^{k + 1}) (b^{k + 1})

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

24. n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.

Solution:

Let P (n): n (n + 1) (n + 5) is a multiple of 3

Let us check for n = 1,

P (1): 1 (1 + 1) (1 + 5)

: 2 × 6

: 12

P (n) is true for n = 1. Where, P (n) is a multiple of 3

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): k (k + 1) (k + 5) is a multiple of 3

: k(k + 1) (k + 5) = 3λ … (i)

We have to prove,

(k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3

(k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ

So,

= (k + 1) [(k + 1) + 1] [(k + 1) + 5]

= (k + 1) (k + 2) [(k + 1) + 5]

= [k (k + 1) + 2(k + 1)] [(k + 5) + 1]

= k (k + 1) (k + 5) + k(k + 1) + 2(k + 1) (k + 5) + 2(k + 1)

= 3λ + k² + k + 2(k² + 6k + 5) + 2k + 2

= 3λ + k² + k + 2k² + 12k + 10 + 2k + 2

= 3λ + 3k² + 15k + 12

= 3(λ + k² + 5k + 4)

= 3μ

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

25. 7²ⁿ + 2^{3n – 3}. 3n – 1 is divisible by 25 for all n ϵ N

Solution:

Let P (n): 7²ⁿ + 2^{3n – 3}. 3n – 1 is divisible by 25

Let us check for n = 1,

P (1): 7² + 2⁰.3⁰

: 49 + 1

: 50

P (n) is true for n = 1. Where, P (n) is divisible by 25

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 7^2k + 2^{3k – 3}. 3k – 1 is divisible by 25.

: 7^2k + 2^{3k – 3}. 3^{k – 1} = 25λ … (i)

We have to prove that

7^{2k + 1} + 2^3k. 3^k is divisible by 25

7^{2k + 2} + 2^3k. 3^k = 25μ

So,

= 7^{2(k + 1)} + 2^3k. 3^k

= 7^2k.7¹ + 2^3k. 3^k

= (25λ – 2^{3k – 3}. 3^{k – 1}) 49 + 2^3k. 3k by using equation (i)

= 25λ. 49 – 2^3k/8. 3^k/3. 49 + 2^3k. 3^k

= 24×25×49λ – 2^3k . 3^k . 49 + 24 . 2^3k.3^k

= 24×25×49λ – 25 . 2^3k. 3^k

= 25(24 . 49λ – 2^3k. 3^k)

= 25μ

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.