Class 11 Mathematics Notes Chapter 4 (Chapter 4) – Examplar Problems (English) Book

Alright class, let's dive into Chapter 4: Principle of Mathematical Induction from your NCERT Exemplar. This is a powerful tool used to prove mathematical statements that are true for all natural numbers. While it might seem a bit abstract initially, it's quite logical and follows a clear structure, making it important for various competitive exams where proof techniques or pattern recognition might be tested indirectly.

Chapter 4: Principle of Mathematical Induction - Detailed Notes

1. What is Mathematical Induction?

Imagine you have an infinite line of dominoes standing upright. If you can ensure two things:
a) The first domino falls.
b) If any domino falls, it knocks over the next one.

Then, you can conclude that all the dominoes will eventually fall.

Mathematical Induction works on a similar principle but for proving mathematical statements, let's call a statement P(n), which depends on a natural number 'n'.

2. The Principle of Mathematical Induction (PMI)

To prove that a statement P(n) is true for all natural numbers n (i.e., n = 1, 2, 3, ...), we need to perform two crucial steps:

• Step 1: Base Case (or Basis Step)

  • Show that the statement P(n) is true for the initial value, usually n = 1. We verify P(1) directly.
  • (Think: Proving the first domino falls).

• Step 2: Inductive Step

  • Part (a): Inductive Hypothesis: Assume that the statement P(n) is true for some arbitrary positive integer 'k'. That is, assume P(k) is true.
    (Think: Assuming some domino 'k' falls).
  • Part (b): Inductive Proof: Using the assumption that P(k) is true, prove that the statement is also true for the next integer, n = k + 1. That is, prove P(k+1) is true.
    (Think: Proving that if domino 'k' falls, it must knock over domino 'k+1').

Conclusion: If both the Base Case (Step 1) and the Inductive Step (Step 2) are established, then by the Principle of Mathematical Induction, the statement P(n) is true for all natural numbers n ∈ N.

3. How to Structure a Proof by Induction

For exams, presenting your proof clearly is vital. Follow these steps:

• Define the Statement: Let P(n) be the statement "..." (write the full statement you need to prove).
• Base Case:
  • "For n = 1,"
  • LHS = ... (calculate the Left Hand Side of the statement for n=1)
  • RHS = ... (calculate the Right Hand Side of the statement for n=1)
  • "Since LHS = RHS, P(1) is true."
• Inductive Step:
  • "Assume P(k) is true for some positive integer k."
  • Write down what P(k) means explicitly: "i.e., ... [Statement with 'k' instead of 'n'] ... (Equation 1)" (This is your Inductive Hypothesis - you will use this).
  • "We need to prove that P(k+1) is true."
  • Write down what P(k+1) means: "i.e., we need to show ... [Statement with 'k+1' instead of 'n'] ..."
  • Now, start with the LHS (or sometimes the more complex side) of P(k+1). Manipulate it algebraically, aiming to use the P(k) statement (Equation 1) from your assumption.
  • Show the steps clearly:
    LHS of P(k+1) = [Expression for P(k+1)]
    = [Break it down, often isolating the terms corresponding to P(k)]
    = [Use Equation 1 (the P(k) assumption) here] + [Remaining term(s)]
    = ... (Algebraic simplification) ...
    = RHS of P(k+1)
  • "Thus, P(k+1) is true whenever P(k) is true."
• Conclusion:
  • "By the Principle of Mathematical Induction, P(n) is true for all n ∈ N."

4. Common Types of Problems Solved Using PMI

• Summation Formulas: Proving formulas for the sum of series.

  • Example: Prove 1 + 2 + 3 + ... + n = n(n+1)/2 for all n ∈ N.
    • P(n): 1 + 2 + ... + n = n(n+1)/2
    • Base Case (n=1): LHS = 1. RHS = 1(1+1)/2 = 1. P(1) is true.
    • Assume P(k): 1 + 2 + ... + k = k(k+1)/2.
    • Prove P(k+1): We need to show 1 + 2 + ... + k + (k+1) = (k+1)(k+1+1)/2 = (k+1)(k+2)/2.
    • LHS = (1 + 2 + ... + k) + (k+1)
      = [k(k+1)/2] + (k+1) (Using P(k))
      = (k+1) [k/2 + 1]
      = (k+1) [(k+2)/2]
      = (k+1)(k+2)/2 = RHS.
    • Conclusion: P(n) is true for all n ∈ N.

• Divisibility Problems: Proving that an expression involving 'n' is divisible by a specific integer.

  • Example: Prove that n³ + 2n is divisible by 3 for all n ∈ N.
    • P(n): n³ + 2n is divisible by 3. (or n³ + 2n = 3m for some integer m)
    • Base Case (n=1): 1³ + 2(1) = 1 + 2 = 3. Since 3 is divisible by 3, P(1) is true.
    • Assume P(k): k³ + 2k is divisible by 3. So, k³ + 2k = 3m for some integer m. (Inductive Hypothesis)
    • Prove P(k+1): We need to show (k+1)³ + 2(k+1) is divisible by 3.
    • (k+1)³ + 2(k+1) = (k³ + 3k² + 3k + 1) + (2k + 2)
      = (k³ + 2k) + 3k² + 3k + 3
      = (k³ + 2k) + 3(k² + k + 1)
      = 3m + 3(k² + k + 1) (Using P(k))
      = 3 [m + k² + k + 1]
    • Since m, k are integers, (m + k² + k + 1) is also an integer. Therefore, (k+1)³ + 2(k+1) is divisible by 3.
    • Conclusion: P(n) is true for all n ∈ N.

• Inequalities: Proving inequalities involving 'n'. These often require careful algebraic manipulation.

  • Example: Prove that 2ⁿ > n for all positive integers n.
    • P(n): 2ⁿ > n
    • Base Case (n=1): LHS = 2¹ = 2. RHS = 1. Since 2 > 1, P(1) is true.
    • Assume P(k): 2ᵏ > k for some positive integer k. (Inductive Hypothesis)
    • Prove P(k+1): We need to show 2ᵏ⁺¹ > k+1.
    • LHS = 2ᵏ⁺¹ = 2 * 2ᵏ

      2 * k (Using P(k))
      = k + k

    • Now, since k is a positive integer, k ≥ 1.
    • So, k + k ≥ k + 1.
    • Combining the inequalities: 2ᵏ⁺¹ > 2k ≥ k+1.
    • Therefore, 2ᵏ⁺¹ > k+1. P(k+1) is true.
    • Conclusion: P(n) is true for all n ∈ N.

5. Important Points for Exams

• Clarity is Key: State P(n), the Base Case, the Inductive Hypothesis, and the goal for P(k+1) clearly.
• Show Your Work: Don't skip algebraic steps, especially in the inductive step where you use P(k) to prove P(k+1).
• Base Case Variations: Sometimes the statement might be true only for n ≥ a, where 'a' is an integer greater than 1 (e.g., n ≥ 4). In such cases, your base case should be n = a (e.g., P(4)). The rest of the process remains the same.
• Don't Assume P(k+1): The most common mistake is assuming P(k+1) is true and working backward. You must start from one side of P(k+1) and derive the other side using the assumption P(k).

Mastering the structure and applying it consistently to different types of problems is the way to succeed with Mathematical Induction.

Multiple Choice Questions (MCQs)

1. The Principle of Mathematical Induction is used to prove statements for:
   a) All real numbers
   b) All integers
   c) All natural numbers
   d) Only n = 1

2. In a proof by induction for a statement P(n), the first step is to:
   a) Assume P(k) is true
   b) Prove P(k+1) is true
   c) Prove P(1) is true
   d) Define P(n)

3. If P(k) is assumed true, the next step in the inductive process is to:
   a) Prove P(1) is true
   b) Prove P(k-1) is true
   c) Prove P(k+1) is true using the assumption P(k)
   d) Assume P(k+1) is true

4. Let P(n) be the statement "n² + n is an even number". What is the base case P(1)?
   a) 1² + 1 = 2, which is even.
   b) k² + k is even.
   c) (k+1)² + (k+1) is even.
   d) 2² + 2 = 6, which is even.

5. Let P(n) be the statement "1 + 3 + 5 + ... + (2n-1) = n²". If P(k) is true, i.e., 1 + 3 + ... + (2k-1) = k², what is the LHS of P(k+1)?
   a) k² + (2k+1)
   b) (k+1)²
   c) 1 + 3 + ... + (2k-1) + (2k+1)
   d) k² + (2(k+1)-1)

6. To prove "7ⁿ - 3ⁿ is divisible by 4" using induction, we assume P(k): "7ᵏ - 3ᵏ = 4m" for some integer m. What expression do we need to show is divisible by 4 for P(k+1)?
   a) 7ᵏ - 3ᵏ
   b) 7ᵏ⁺¹ - 3ᵏ⁺¹
   c) 4(k+1)
   d) 7(k+1) - 3(k+1)

7. Consider the statement P(n): "3ⁿ > n+2". For which starting value of n should the base case be checked if we want to prove it for n ≥ 2?
   a) n = 1
   b) n = 0
   c) n = 2
   d) n = 3

8. In proving 1² + 2² + ... + n² = n(n+1)(2n+1)/6, the inductive step requires showing P(k+1) is true assuming P(k) is true. The expression 1² + 2² + ... + k² + (k+1)² simplifies using P(k) to:
   a) k(k+1)(2k+1)/6 + (k+1)
   b) k(k+1)(2k+1)/6 + (k+1)²
   c) (k+1)(k+2)(2k+3)/6
   d) k² + (k+1)²

9. If P(1) is true and P(k+1) is true whenever P(k) is true, then P(n) is true for:
   a) Only n=1 and n=k+1
   b) All n ∈ N
   c) Only n=1, 2, 3
   d) All n > 1

10. A common error in proving P(k+1) in the inductive step is:
    a) Forgetting the base case P(1).
    b) Assuming P(k+1) is true initially and working backward.
    c) Making algebraic errors while simplifying the expression for P(k+1).
    d) All of the above.

Answer Key for MCQs:

1. c
2. c (Although defining P(n) is the very first conceptual step, proving P(1) is the first action in the proof structure).
3. c
4. a
5. c (Option 'a' and 'd' represent the expression after using the P(k) assumption, not the initial LHS of P(k+1)).
6. b
7. c
8. b
9. b
10. d (All are common issues or errors).