3. 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.
4.
5.
6.
7. Prove for ≥ 1
1 x 1! + 2 x 2! + 3 x 3! + ... + n x n! = (n +
1)! - 1
This could be also written by using ∑ notation
8.
9. Proof
Base case: n + 1
The left hand side is 1x1! The right hand side is
2! - 1. They are equal.
Inductive hypothesis: Suppose this holds
10.
11. 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
12.
13. We can picture each proposition as a domino:
P (k)
39. Then, ∀ n P (n) is true.
Use induction to prove that the sum of the first n
odd integers is
Base case (n=1): the sum of the first 1 odd integer
is
Assume p (k): the sum of the first k odd integers
is
Yeah!
Where,
1 + 3 + … + (2k+1) = =1
Prove that
1 + 3 + … + + (2k-1) + (2k+1) =
1 + 3 + … + + (2k-1) + (2k+1) =
=
40.
41. Prove that: 1 x 1! + 2 x 2! + … + n x n! = (n+1)! -1, ∀ n
Base case (n=1): 1 x 1! = (1 x 1)! -1?
Assume P (k 1 x 1! + 2 x 2! + … + k x k! = (k+1)! -1
Prove 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
