1. UNIVERSITI PENDIDKAN SULTAN IDRIS PREPARED BY : MOHAMAD AL FAIZ BIN SELAMAT
2.  Principle of Induction. To prove that P (n) is true for all positive integers n, where P (n) is a propositional function. A proof by mathematical induction has two parts: Basic step: We verify that P (1) is true. Inductive step: We show that the conditional proposition ∀ k ∈ , P (k) ⇒ P (k + 1) is true.
3. Prove for ≥ 1 1 x 1! + 2 x 2! + 3 x 3! + ... + n x n! = (n + 1)! - 1This could be also written by using ∑ notation
4. ProofBase case: n + 1The left hand side is 1x1! The right hand side is2! - 1. They are equal.Inductive hypothesis: Suppose this holds
5.  We need to prove Consider the left hand side= (n+1)! – 1 + (n+1) x (n+1)= (n+1)! (1+n+1) -1 = (n+2)! -1
6.  We can picture each proposition as a domino: P (k)
7. P (0) P (1) P (2) P (k) P (k+1) …..
8. P (0) P (1) P (2) P (k) P (k+1) …..
9. P (1) P (2) P (k) P (k+1) …..
10. P (2) P (k) P (k+1) …..
11. P (k) P (k+1)…..
12. P (k) P (k+1)…..
13. P (k+1)…..
14. Then, ∀ n P (n) is true.Use induction to prove that the sum of the first nodd integers isBase case (n=1): the sum of the first 1 odd integerisAssume p (k): the sum of the first k odd integersis Yeah!Where, 1 + 3 + … + (2k+1) = =1Prove that 1 + 3 + … + + (2k-1) + (2k+1) = 1 + 3 + … + + (2k-1) + (2k+1) = =
15. Prove that: 1 x 1! + 2 x 2! + … + n x n! = (n+1)! -1, ∀ nBase case (n=1): 1 x 1! = (1 x 1)! -1?Assume P (k 1 x 1! + 2 x 2! + … + k x k! = (k+1)! -1Prove that: x 1! + 2 x 2! + … + k x k! + (k+1) (k+1)! = (k+2)! -1 1 x 1! + 2 x 2! + … + k x k! + (k+1) (k+1)! = (k+1)! -1+ (k+1) (k+1) = (1 + (k+1)) (k+1)! – 1 = (k+2) (k+1)! -1 = (k+2)! - 1 1 x 1! = 1 2! – 1 = 1
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