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Mathematical induction
Mathematical induction is a method of proof that can be used to prove that a statement is true for all positive integers. It involves two steps: 1) proving the statement is true for the base case, usually n = 1, and 2) assuming the statement is true for an integer k and using this to prove the statement is true for k + 1. Examples are provided to demonstrate how to use mathematical induction to prove statements such as the sum of the first n positive integers equalling n(n+1)/2 and that 7n - 1 is divisible by 6 for all positive integers n.
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Mathematical induction
- 1. MATHEMATICAL INDUCTION PROVING BYMATHEMATICAL INDUCTION PreCalculus
- 2. THE PRINCIPLE OFMATHEMATICAL INDUCTION •Let Sn be a statement for each positive integer n. suppose that the following condition hold: •1. S 𝟏 is true. •2. If Sk is true, then Sk+1 should be true, where k is any positive integer. •Therefore, Sn is true for all positive integers n.
- 3. ILLUSTRATIVE EXAMPLES: •Example 1:Using mathematical induction, prove that 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏 = 𝒏 𝒏+𝟏 𝟐 for all positive integers n.
- 4. PMI Step 1: n= 1 𝟏 = 𝟏 𝟏+𝟏 𝟐 = 𝟐 𝟐 = 1 • Example 1: Using mathematical induction, prove that 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏 = 𝒏 𝒏+𝟏 𝟐 for all positive integers n.
- 5. PMI Step 2: forn = k 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒌 = 𝒌 𝒌+𝟏 𝟐 and for n = k + 1 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒌 + (𝒌 + 𝟏) = (𝒌 + 𝟏) 𝒌 + 𝟐 𝟐 • Example 1: Using mathematical induction, prove that 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏 = 𝒏 𝒏+𝟏 𝟐 for all positive integers n.
- 6. PMI and for n= k + 1 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒌 + (𝒌 + 𝟏) = (𝒌 + 𝟏) 𝒌 + 𝟐 𝟐 𝒌 𝒌+𝟏 𝟐 + (k + 1) = (𝒌+𝟏) 𝒌+𝟐 𝟐 • Example 1: Using mathematical induction, prove that 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏 = 𝒏 𝒏+𝟏 𝟐 for all positive integers n.
- 7. PMI 𝒌 𝒌+𝟏 𝟐 + (k+ 1) = (𝒌+𝟏) 𝒌+𝟐 𝟐 𝒌 𝒌+𝟏 +𝟐(𝒌+𝟏) 𝟐 = 𝒌+𝟏 (𝒌+𝟐) 𝟐 = (𝒌+𝟏) 𝒌+𝟐 𝟐 • Example 1: Using mathematical induction, prove that 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏 = 𝒏 𝒏+𝟏 𝟐 for all positive integers n.
- 8. ILLUSTRATIVE EXAMPLES •Example 2:Prove that 𝒊=𝟏 𝒏 𝟑𝒊 = 𝟑𝒏(𝒏 + 𝟏) 𝟐 .
- 9. EXAMPLE 3 Using mathematicalinduction, prove that 𝟏 𝟐 + 𝟐 𝟐 + 𝟑 𝟐 + ⋯ + 𝒏 𝟐 = 𝒏 𝒏+𝟏 (𝟐𝒏+𝟏) 𝟔 for all positive integers n.
- 10. EXAMPLE 4 •Use mathematicalinduction to prove that, for every positive integer n, 7n – 1 is divisible by 6.
- 11. EXAMPLE 4 •Part 1 71−1 = 6 = 6 · 1 71 − 1 is divisible by 6.
- 12. EXAMPLE 4 Part 2 Assume:7k − 1 is divisible by 6. To show: 7k+1 − 1 is divisible by 6.
- 13. Part 2 To show:7k+1 − 1 is divisible by 6. 7k+1 − 1 = 7 · 7k − 1 = 6 · 7k + 7k −1 = 6 · 7k + (7k − 1)
- 14. To show: 7k+1− 1 is divisible by 6. By definition of divisibility, 6 · 7k is divisible by 6. Also, by the hypothesis (assumption), 7k − 1 is divisible by 6. Hence, their sum (which is equal to 7k+1 − 1) is also divisible by 6.
- 15. EXAMPLE 5 •Use mathematicalinduction to prove that, for every nonnegative integer n, 3n-n+3 is divisible by 3.
- 16. EXAMPLE 6 •Use mathematicalinduction to prove that 2n > 2n for every integer n ≥ 3.
- 17. Reference: • Aoanan, GraceO. et. al. (2018) General Mathematics for Senior HS, C&E Publishing, Inc. • Garces, Ian June L. et. al. (2016) Precalculus “Teaching Guide for Senior High School, Commission on Higher
- 18.