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Chapter 4 Sequences and Mathematical Induction

4.2 Mathematical Induction

A recent method for proving mathematical arguments. De Morgan is credited with its discovery and and name. The validity of proof by mathematical induction is taken as a axiom. – an axiom or postulate is a proposition that is not proved or demonstrated but considered to be either self-evident, or subject to necessary decision. (Wikipedia) self-evidentdecision

Principle of Mathematical Induction Let P(n) be a property that is defined for integers n, and let a be a fixed integer. Suppose the following two statements are true: 1.P(a) is true. 2.For all integers k ≥ a, if P(k) is true then P(k+1) is true. Then the statement, for all integers n ≥ a, P(a) is true

Example For all integers n ≥ 8, n cents can be obtained using 3 cents and 5 cents. or, for all integers n ≥ 8, P(n) is true, where P(n) is the sentence “n cents can be obtained using 3 cents and 5 cents.”

Example This can be proven by exhaustion if we can continue to fill in the table up to \$1.00. The table shows how to obtain k cents using 3 and 5 coins. We must show how to obtain (k+1) cents. Two cases: – k: 3 + 5, k+1: ? replace 5 with 3 + 3 – k: 3 + 3 + 3, k+1: ? replace 3+3+3 with 5+5

Method of Proof Mathematical Induction Statement: “For all integers n≥a, a property P(n) is true.” (basis step) Show that the property is true for n = a. (inductive step) Show that for all integers k≥a, if the property is true for n=k then it is true for n=k+1. – (inductive hypothesis) suppose that the property is true for n=k, where k is any particular but arbitrarily chosen integer with k≥a. – then, show that the property is true for n = k+1

Example Coins Revisited Proposition 4.2.1: Let P(n) be the property “n cents can be obtained using 3 and 5 cent coins.” Then P(n) is true for all integers n≥8. – Proof: – Show that the property is true for n=8: The property is true b/c 8=3+5.

Example cont. – Show that for all integers k≥8, if the property true for n=k, then property true for n=k+1 (inductive hypothesis) Suppose k cents can be obtained using 3 and 5 cent coins for k≥8. Must show (k+1) cents can be obtained from 3 & 5 coin. – Case (3,5 coin): k+1 can be obtained by replacing the 5 coin with two 3 cent coins. This increments the value by 1 (3+3=6) replaces the 5 cent coin. – Case (3,3,3 coin): k+1 can be obtained by replacing the three 3 coins with two 5 coins. k=b+3+3+3=b+9 and k+1=b+9+1=b+5+5=b+10

Example Formula Prove with mathematical induction Identify P(n) Basis step

Example Formula cont. Inductive step – assume P(k) is true, k>=1 – show that P(k+1) is true by subing k+1 for n – show that left side 1+2+…k+1 = right side (k+1)(k+2)/2 – 1+2+…+k+1 = (1+2+…+k) + k+1 – sub from inductive hypothesis:

Theorem 4.2.2 Sum of the First n Integers – For all integers n≥1,

Example Sum of the First n Integers – Find 2+4+6+…+500 Get in form of Theorem 4.2.2 (1+2+…+n) factor out 2: 2(1+2+3+…+250) sum = 2( n(n+1)/2 ), n = 250 sum = 2( 250(250+1)/2 ) = 62,750 – Find 5+6+7+8+…+50 add first 4 terms 1+2+3+4 to problem then subtract back out after computation with 4.2.2 1+2+3+4+5+6+7+…+50 – (1+2+3+4) (50 (50+1)/ 2) – 10 =1,265

Sum of Geometric Sequence Prove that, for all integers n≥0 and all real numbers r except 1. P(n): Basis: Inductive: (n=k) Inductive Hypothesis: (n=k+1)

Theorem 4.2.3 Sum of Geometric Sequences – For any real number r except 1, and any integer n≥0,

Examples Sum of Geometric Sequences – Find 1 + 3 + 3 2 + … + 3 m-2 Sequence is in geometric series, apply 4.2.3 directly – Find 3 2 + 3 3 +… + 3 m rearrange into proper geometric sequence by factoring out 3 2 from sequence 3 2 (1 + 3 + 3 2 + … + 3 m-2 ) =

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