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Use mathematical induction to prove the following statement. n+1 : 2, (¹ - 2 ) (¹ - ¹3 ) --- (¹ - 22 ) -^2¹ Proof (by mathematical induction): Let the property P(n) be the equation (₁-21) (¹ - 1) (¹ - 1) - We will show that P(n) is true for every integer n 2 For every integer n 2 2, Show that P Show that for each integer k is true: Before simplification, the left-hand side of Pis (¹-2)(¹-3) (¹ - 1) and the right-hand side of P(k) is We must show that |- (¹ - 2 ) ) ( ¹ - 3 - ) ( ¹ - - - - - - - (¹ - 1- n+1 2n if P(k) is true, then P 2k [The inductive hypothesis is that the two sides of P(k) are equal.] PC When the left- and right-hand sides of P 1 +1 Before simplification, the right-hand side of P and the next-to-last factor in the left-hand side is (1-21)- (¹ - 2 ) (¹ - ² ) (¹ - - ) --- (1 - 2 ) ( ¹ - ( 1 is true. In other words, we must show that the left and right-hand sides of P After substitution from the inductive hypothesis, the left-hand side of P I).(+-+ T)-( 1 2k -O). 2k(k+1) 1- 2(k+1) So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is and the right-hand side is becomes 1 +1. After simplification, both sides can be shown to equal 2.2 are simplified, both can be shown to equal are equal. The left-hand side of . Thus, P is true.

Use mathematical induction to prove the following statement. n+1 : 2, (¹ - 2 ) (¹ - ¹3 ) --- (¹ - 22 ) -^2¹ Proof (by mathematical induction): Let the property P(n) be the equation (₁-21) (¹ - 1) (¹ - 1) - We will show that P(n) is true for every integer n 2 For every integer n 2 2, Show that P Show that for each integer k is true: Before simplification, the left-hand side of Pis (¹-2)(¹-3) (¹ - 1) and the right-hand side of P(k) is We must show that |- (¹ - 2 ) ) ( ¹ - 3 - ) ( ¹ - - - - - - - (¹ - 1- n+1 2n if P(k) is true, then P 2k [The inductive hypothesis is that the two sides of P(k) are equal.] PC When the left- and right-hand sides of P 1 +1 Before simplification, the right-hand side of P and the next-to-last factor in the left-hand side is (1-21)- (¹ - 2 ) (¹ - ² ) (¹ - - ) --- (1 - 2 ) ( ¹ - ( 1 is true. In other words, we must show that the left and right-hand sides of P After substitution from the inductive hypothesis, the left-hand side of P I).(+-+ T)-( 1 2k -O). 2k(k+1) 1- 2(k+1) So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is and the right-hand side is becomes 1 +1. After simplification, both sides can be shown to equal 2.2 are simplified, both can be shown to equal are equal. The left-hand side of . Thus, P is true.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter2: The Integers
Section2.2: Mathematical Induction
Problem 46E: Use generalized induction and Exercise 43 to prove that n22n for all integers n5. (In connection...
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Use mathematical induction to prove the following statement. 1 n+ 22, (1 - 12/2) (¹ - 1) (¹ - 22 ) - 02/1 1 1 = 2n Proof (by mathematical induction): Let the property P(n) be the equation 1 (¹₁ - 1212) (¹ - 3 2 ) - ( ¹₁ - 12/1 ) - 1 1 1 3² We will show that P(n) is true for every integer n 2 Show that P For every integer n ≥ 2, 1- Show that for each integer k ≥ (¹ - 2 ) ( ¹ - - - ) ( ¹ - - - ) . · ( ¹ - 1 1 3² and the right-hand side of P(k) is is true: Before simplification, the left-hand side of P We must show that P 1 (¹ - 1) (¹ - ¹) (¹ - 4- ) --- (¹ 1 1 1 - 22 2k , if P(k) is true, then P [The inductive hypothesis is that the two sides of P(k) are equal.] n+ 1 2n 2k Hence, P 1 When the left and right-hand sides of P + 1 and the next-to-last factor in the left-hand side is 1- is (1-1/2) 1 (¹ - 2 ) ( ¹ - 3 - ) ( ¹ - - - - - - - (1 - 2 ) ( ¹ 1- 1 1 121) (₁ - ( 1- 2² 4² k² Before simplification, the right-hand side of P 1 is true. In other words, we must show that the left- and right-hand sides of P After substitution from the inductive hypothesis, the left-hand side of P 1 I)·(²-²)³²) - ( 1 2k 2k(k+1) Y) is So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is is 0(1-11 2k is true, which completes the inductive step. 2 (K + 1) +1 and the right-hand side is is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is becomes 1 are simplified, both can be shown to equal + 1 2.2 After simplification, both sides can be shown to equal are equal. The left-hand side of P is Thus, P is true.
Transcribed Image Text:Use mathematical induction to prove the following statement. 1 n+ 22, (1 - 12/2) (¹ - 1) (¹ - 22 ) - 02/1 1 1 = 2n Proof (by mathematical induction): Let the property P(n) be the equation 1 (¹₁ - 1212) (¹ - 3 2 ) - ( ¹₁ - 12/1 ) - 1 1 1 3² We will show that P(n) is true for every integer n 2 Show that P For every integer n ≥ 2, 1- Show that for each integer k ≥ (¹ - 2 ) ( ¹ - - - ) ( ¹ - - - ) . · ( ¹ - 1 1 3² and the right-hand side of P(k) is is true: Before simplification, the left-hand side of P We must show that P 1 (¹ - 1) (¹ - ¹) (¹ - 4- ) --- (¹ 1 1 1 - 22 2k , if P(k) is true, then P [The inductive hypothesis is that the two sides of P(k) are equal.] n+ 1 2n 2k Hence, P 1 When the left and right-hand sides of P + 1 and the next-to-last factor in the left-hand side is 1- is (1-1/2) 1 (¹ - 2 ) ( ¹ - 3 - ) ( ¹ - - - - - - - (1 - 2 ) ( ¹ 1- 1 1 121) (₁ - ( 1- 2² 4² k² Before simplification, the right-hand side of P 1 is true. In other words, we must show that the left- and right-hand sides of P After substitution from the inductive hypothesis, the left-hand side of P 1 I)·(²-²)³²) - ( 1 2k 2k(k+1) Y) is So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is is 0(1-11 2k is true, which completes the inductive step. 2 (K + 1) +1 and the right-hand side is is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is becomes 1 are simplified, both can be shown to equal + 1 2.2 After simplification, both sides can be shown to equal are equal. The left-hand side of P is Thus, P is true.
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