RD Sharma Class 11 Solutions for Chapter 12: Mathematical Induction

RD Sharma Class 11 Solutions for Chapter 12: Mathematical Induction are an indispensable resource for students aiming to master advanced proof techniques in the Class 11 Maths syllabus. This chapter introduces students to the Principle of Mathematical Induction, providing a powerful and systematic method for proving various mathematical statements, including summation formulas, divisibility properties, and inequalities.

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

These RD Sharma Solutions for Class 11 are made by experts at Infinity Learn to be your go-to guide. They help you quickly clear up any confusion and aim for high scores. No matter if you're studying for CBSE or other state boards, these step-by-step solutions make sure you truly get the hang of Mathematical Induction. This PDF download for RD Sharma Solutions for Class 11 Maths Chapter 12 Mathematical Induction is perfect for offline study.

This chapter covers key ideas like:

• The Principle of Mathematical Induction itself.
• Proving Summation Formulas (like the sum of natural numbers).
• Proving Divisibility Statements.
• Proving Inequalities.
• How to use Mathematical Induction in other math problems.

These topics are the groundwork for tougher math concepts later on, and they're really important for higher studies in math and computer science.

RD Sharma Solutions for Class 11 Maths Chapter 12: Step by Step Solution

Download your free PDF to get a step-by-step guide to Mathematical Induction for competitive exams and do better in your Class 11 Maths studies with trusted RD Sharma Solutions. You'll master how to tackle summation formulas, divisibility proofs, and inequalities using this powerful proof technique. Learn the crucial basis step and inductive step and avoid common mistakes in Mathematical Induction proofs.

RD Sharma Solutions for Class 11 Mathematical Induction

1. Verify P(3) for P(n): n(n+1) is even

Solution:

For n = 3:

P(3): 3(3+1) = 3 × 4 = 12

Since 12 is divisible by 2, P(3) is true.

Hence, P(3) is verified.

2. Prove 2^n > n for all n ∈ ℕ

Solution:

Step 1: Base case (n = 1)

2^1 = 2 > 1 ✓

Step 2: Inductive hypothesis

Assume 2^k > k for some k ∈ ℕ

Step 3: Inductive step

We need to prove 2^(k+1) > k+1

2^(k+1) = 2 × 2^k

Since 2^k > k (by hypothesis), we have:

2^(k+1) = 2 × 2^k > 2k

For k ≥ 1, we have 2k ≥ k+1

Therefore, 2^(k+1) > k+1

By mathematical induction, 2^n > n for all n ∈ ℕ

3. Show 1 + 3 + 5 + ... + (2n-1) = n²

Solution:

Step 1: Base case (n = 1)

LHS = 2(1) - 1 = 1

RHS = 1² = 1

LHS = RHS ✓

Step 2: Inductive hypothesis

Assume 1 + 3 + 5 + ... + (2k-1) = k²

Step 3: Inductive step

We need to prove: 1 + 3 + 5 + ... + (2k-1) + (2(k+1)-1) = (k+1)²

LHS = k² + (2k+1) = k² + 2k + 1 = (k+1)²

Therefore, the formula holds for all n ∈ ℕ

4. Prove 1² + 2² + ... + n² = n(n+1)(2n+1)/6

Solution:

Step 1: Base case (n = 1)

LHS = 1² = 1

RHS = 1(2)(3)/6 = 1

LHS = RHS ✓

Step 2: Inductive hypothesis

Assume 1² + 2² + ... + k² = k(k+1)(2k+1)/6

Step 3: Inductive step

We need to prove: 1² + 2² + ... + k² + (k+1)² = (k+1)(k+2)(2k+3)/6

LHS = k(k+1)(2k+1)/6 + (k+1)²

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

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

= (k+1)[2k² + k + 6k + 6]/6

= (k+1)[2k² + 7k + 6]/6

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

Therefore, the formula holds for all n ∈ ℕ

5. Demonstrate 1³ + 2³ + ... + n³ = [n(n+1)/2]²

Solution:

Step 1: Base case (n = 1)

LHS = 1³ = 1

RHS = [1(2)/2]² = 1² = 1

LHS = RHS ✓

Step 2: Inductive hypothesis

Assume 1³ + 2³ + ... + k³ = [k(k+1)/2]²

Step 3: Inductive step

We need to prove: 1³ + 2³ + ... + k³ + (k+1)³ = [(k+1)(k+2)/2]²

LHS = [k(k+1)/2]² + (k+1)³

= [k²(k+1)²/4] + (k+1)³

= (k+1)²[k²/4 + (k+1)]

= (k+1)²[k² + 4k + 4]/4

= (k+1)²(k+2)²/4

= [(k+1)(k+2)/2]²

Therefore, the formula holds for all n ∈ ℕ

6. Verify 1·2 + 2·3 + ... + n(n+1) = n(n+1)(n+2)/3

Solution:

Step 1: Base case (n = 1)

LHS = 1·2 = 2

RHS = 1(2)(3)/3 = 2

LHS = RHS ✓

Step 2: Inductive hypothesis

Assume 1·2 + 2·3 + ... + k(k+1) = k(k+1)(k+2)/3

Step 3: Inductive step

We need to prove: 1·2 + 2·3 + ... + k(k+1) + (k+1)(k+2) = (k+1)(k+2)(k+3)/3

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

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

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

Therefore, the formula holds for all n ∈ ℕ

7. Prove 1·3 + 3·5 + ... + (2n-1)(2n+1) = n(4n²+6n-1)/3

Solution:

Step 1: Base case (n = 1)

LHS = 1·3 = 3

RHS = 1(4+6-1)/3 = 9/3 = 3

LHS = RHS ✓

Step 2: Inductive hypothesis

Assume 1·3 + 3·5 + ... + (2k-1)(2k+1) = k(4k²+6k-1)/3

Step 3: Inductive step

We need to prove the formula for n = k+1

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

= k(4k²+6k-1)/3 + (2k+1)(2k+3)

= k(4k²+6k-1)/3 + (4k²+8k+3)

= [k(4k²+6k-1) + 3(4k²+8k+3)]/3

= [4k³+6k²-k + 12k²+24k+9]/3

= [4k³+18k²+23k+9]/3

= (k+1)[4(k+1)²+6(k+1)-1]/3

Therefore, the formula holds for all n ∈ ℕ

8. Show 7ⁿ - 3ⁿ is divisible by 4 for all n ∈ ℕ

Solution:

Step 1: Base case (n = 1)

7¹ - 3¹ = 7 - 3 = 4, which is divisible by 4 ✓

Step 2: Inductive hypothesis

Assume 7ᵏ - 3ᵏ is divisible by 4

Step 3: Inductive step

We need to prove 7^(k+1) - 3^(k+1) is divisible by 4

7^(k+1) - 3^(k+1) = 7·7ᵏ - 3·3ᵏ

= 7·7ᵏ - 7·3ᵏ + 7·3ᵏ - 3·3ᵏ

= 7(7ᵏ - 3ᵏ) + 3ᵏ(7 - 3)

= 7(7ᵏ - 3ᵏ) + 4·3ᵏ

Since 7ᵏ - 3ᵏ is divisible by 4 (hypothesis) and 4·3ᵏ is divisible by 4,

their sum is divisible by 4.

Therefore, 7ⁿ - 3ⁿ is divisible by 4 for all n ∈ ℕ

9. Prove 11ⁿ - 6 is divisible by 5 for all n ≥ 1

Solution:

Step 1: Base case (n = 1)

11¹ - 6 = 11 - 6 = 5, which is divisible by 5 ✓

Step 2: Inductive hypothesis

Assume 11ᵏ - 6 is divisible by 5

Step 3: Inductive step

We need to prove 11^(k+1) - 6 is divisible by 5

11^(k+1) - 6 = 11·11ᵏ - 6

= 11·11ᵏ - 11·6 + 11·6 - 6

= 11(11ᵏ - 6) + 6(11 - 1)

= 11(11ᵏ - 6) + 60

Since 11ᵏ - 6 is divisible by 5 (hypothesis) and 60 = 12×5 is divisible by 5,

their sum is divisible by 5.

Therefore, 11ⁿ - 6 is divisible by 5 for all n ≥ 1

10. Demonstrate 3^(2n) - 1 is divisible by 8 for all natural n

Solution:

Step 1: Base case (n = 1)

3^(2·1) - 1 = 3² - 1 = 9 - 1 = 8, which is divisible by 8 ✓

Step 2: Inductive hypothesis

Assume 3^(2k) - 1 is divisible by 8

Step 3: Inductive step

We need to prove 3^(2(k+1)) - 1 is divisible by 8

3^(2(k+1)) - 1 = 3^(2k+2) - 1 = 3²·3^(2k) - 1 = 9·3^(2k) - 1

= 9·3^(2k) - 9 + 9 - 1

= 9(3^(2k) - 1) + 8

Since 3^(2k) - 1 is divisible by 8 (hypothesis) and 8 is divisible by 8,

their sum is divisible by 8.

Therefore, 3^(2n) - 1 is divisible by 8 for all natural n

11. Verify n(n+1)(n+2) is divisible by 6 for all n ∈ ℕ

Solution:

Step 1: Base case (n = 1)

1(2)(3) = 6, which is divisible by 6 ✓

Step 2: Inductive hypothesis

Assume k(k+1)(k+2) is divisible by 6

Step 3: Inductive step

We need to prove (k+1)(k+2)(k+3) is divisible by 6

Since 6 = 2×3, we need to show divisibility by both 2 and 3

Divisibility by 2: Among three consecutive integers (k+1), (k+2), (k+3), at least one is even

Divisibility by 3: Among three consecutive integers, exactly one is divisible by 3

Therefore, (k+1)(k+2)(k+3) is divisible by 6

Hence, n(n+1)(n+2) is divisible by 6 for all n ∈ ℕ

12. Prove 5ⁿ + 2·11ⁿ is divisible by 3 for all n ≥ 1

Solution:

Step 1: Base case (n = 1)

5¹ + 2·11¹ = 5 + 22 = 27, which is divisible by 3 ✓

Step 2: Inductive hypothesis

Assume 5ᵏ + 2·11ᵏ is divisible by 3

Step 3: Inductive step

We need to prove 5^(k+1) + 2·11^(k+1) is divisible by 3

5^(k+1) + 2·11^(k+1) = 5·5ᵏ + 2·11·11ᵏ

= 5·5ᵏ + 22·11ᵏ

= 5·5ᵏ + 2·11ᵏ + 20·11ᵏ

= (5ᵏ + 2·11ᵏ) + 5ᵏ(5-1) + 20·11ᵏ

= (5ᵏ + 2·11ᵏ) + 4·5ᵏ + 20·11ᵏ

Since 5ᵏ + 2·11ᵏ is divisible by 3 (hypothesis), and we need to check if 4·5ᵏ + 20·11ᵏ is divisible by 3

Note: 5 ≡ 2 (mod 3) and 11 ≡ 2 (mod 3)

So 5ᵏ ≡ 2ᵏ (mod 3) and 11ᵏ ≡ 2ᵏ (mod 3)

Therefore, 5ᵏ + 2·11ᵏ ≡ 2ᵏ + 2·2ᵏ = 3·2ᵏ ≡ 0 (mod 3)

Hence, 5ⁿ + 2·11ⁿ is divisible by 3 for all n ≥ 1

13. Prove 2ⁿ > 2n + 1 for n ≥ 3

Solution:

Step 1: Base case (n = 3)

2³ = 8 > 2(3) + 1 = 7 ✓

Step 2: Inductive hypothesis

Assume 2ᵏ > 2k + 1 for some k ≥ 3

Step 3: Inductive step

We need to prove 2^(k+1) > 2(k+1) + 1 = 2k + 3

2^(k+1) = 2·2ᵏ > 2(2k + 1) = 4k + 2

We need to show 4k + 2 > 2k + 3, i.e., 2k > 1

Since k ≥ 3, we have 2k ≥ 6 > 1

Therefore, 2^(k+1) > 2(k+1) + 1

Hence, 2ⁿ > 2n + 1 for all n ≥ 3

14. Show n⁵/5 + n³/3 + 7n/15 is a natural number for all n ∈ ℕ

Solution:

Let f(n) = n⁵/5 + n³/3 + 7n/15 = (3n⁵ + 5n³ + 7n)/15

Step 1: Base case (n = 1)

f(1) = (3 + 5 + 7)/15 = 15/15 = 1 ∈ ℕ ✓

Step 2: Inductive hypothesis

Assume f(k) is a natural number

Step 3: Inductive step

f(k+1) - f(k) = [(k+1)⁵ - k⁵]/5 + [(k+1)³ - k³]/3 + 7/15

Using binomial expansion:

(k+1)⁵ - k⁵ = 5k⁴ + 10k³ + 10k² + 5k + 1

(k+1)³ - k³ = 3k² + 3k + 1

f(k+1) - f(k) = (5k⁴ + 10k³ + 10k² + 5k + 1)/5 + (3k² + 3k + 1)/3 + 7/15

= k⁴ + 2k³ + 2k² + k + 1/5 + k² + k + 1/3 + 7/15

= k⁴ + 2k³ + 3k² + 2k + (3 + 5 + 7)/15

= k⁴ + 2k³ + 3k² + 2k + 1

This is clearly a natural number, so f(k+1) is also a natural number

Therefore, n⁵/5 + n³/3 + 7n/15 is a natural number for all n ∈ ℕ

15. Prove 1/(1·2) + 1/(2·3) + ... + 1/[n(n+1)] = n/(n+1)

Solution:

Step 1: Base case (n = 1)

LHS = 1/(1·2) = 1/2

RHS = 1/(1+1) = 1/2

LHS = RHS ✓

Step 2: Inductive hypothesis

Assume 1/(1·2) + 1/(2·3) + ... + 1/[k(k+1)] = k/(k+1)

Step 3: Inductive step

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

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

= (k² + 2k + 1)/[(k+1)(k+2)]

= (k+1)²/[(k+1)(k+2)]

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

Therefore, the formula holds for all n ∈ ℕ

16. Verify 1/(1·3) + 1/(3·5) + ... + 1/[(2n-1)(2n+1)] = n/(2n+1)

Solution:

Step 1: Base case (n = 1)

LHS = 1/(1·3) = 1/3

RHS = 1/(2+1) = 1/3

LHS = RHS ✓

Step 2: Inductive hypothesis

Assume 1/(1·3) + 1/(3·5) + ... + 1/[(2k-1)(2k+1)] = k/(2k+1)

Step 3: Inductive step

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

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

= (2k² + 3k + 1)/[(2k+1)(2k+3)]

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

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

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

Therefore, the formula holds for all n ∈ ℕ

17. Show 1² - 2² + 3² - ... + (-1)^(n-1)n² = (-1)^(n-1)n(n+1)/2

Solution:

Step 1: Base case (n = 1)

LHS = (-1)⁰·1² = 1

RHS = (-1)⁰·1·2/2 = 1

LHS = RHS ✓

Step 2: Base case (n = 2)

LHS = 1² - 2² = 1 - 4 = -3

RHS = (-1)¹·2·3/2 = -3

LHS = RHS ✓

Step 3: Inductive hypothesis

Assume 1² - 2² + 3² - ... + (-1)^(k-1)k² = (-1)^(k-1)k(k+1)/2

Step 4: Inductive step

LHS = (-1)^(k-1)k(k+1)/2 + (-1)ᵏ(k+1)²

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

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

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

= (-1)ᵏ(k+1)(k+2)/2

Therefore, the formula holds for all n ∈ ℕ

18. Prove 1 + 1/(1+2) + 1/(1+2+3) + ... + 1/(1+2+...+n) = 2n/(n+1)

Solution:

Note: 1+2+...+k = k(k+1)/2

Step 1: Base case (n = 1)

LHS = 1 = 1

RHS = 2·1/(1+1) = 2/2 = 1

LHS = RHS ✓

Step 2: Inductive hypothesis

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

Step 3: Inductive step

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

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

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

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

= [2k² + 4k + 2]/[(k+1)(k+2)]

= 2(k² + 2k + 1)/[(k+1)(k+2)]

= 2(k+1)²/[(k+1)(k+2)]

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

Therefore, the formula holds for all n ∈ ℕ

19. Prove the number of subsets of a set with n elements is 2ⁿ

Solution:

Step 1: Base case (n = 1)

A set with 1 element {a} has subsets: ∅, {a}

Number of subsets = 2 = 2¹ ✓

Step 2: Inductive hypothesis

Assume a set with k elements has 2ᵏ subsets

Step 3: Inductive step

Consider a set S with (k+1) elements. Let S = T ∪ {x} where T has k elements

Subsets of S are of two types:

1. Subsets not containing x: These are exactly the subsets of T, so there are 2ᵏ of them

2. Subsets containing x: These are of the form A ∪ {x} where A is a subset of T, so there are 2ᵏ of them

Total subsets = 2ᵏ + 2ᵏ = 2^(k+1)

Therefore, a set with n elements has 2ⁿ subsets

20. Show 1 + 5 + 9 + ... + (4n-3) = n(2n-1)

Solution:

Step 1: Base case (n = 1)

LHS = 4(1) - 3 = 1

RHS = 1(2-1) = 1

LHS = RHS ✓

Step 2: Inductive hypothesis

Assume 1 + 5 + 9 + ... + (4k-3) = k(2k-1)

Step 3: Inductive step

We need to prove: 1 + 5 + 9 + ... + (4k-3) + (4(k+1)-3) = (k+1)(2(k+1)-1)

LHS = k(2k-1) + (4k+4-3)

= k(2k-1) + (4k+1)

= 2k² - k + 4k + 1

= 2k² + 3k + 1

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

= (k+1)(2(k+1)-1)

Therefore, the formula holds for all n ∈ ℕ