Use mathematical induction to prove the following statement. n+1 ≥ 2₁ (1-2) (₁-3) (1 - ²2 ) -^21 Proof (by mathematical induction): Let the property P(n) be the equation (¹-2)(¹-3)... - 2²2) = ² + 1. 20 We will show that P(n) is true for every integer n z. For every integer n 2 Show that P) is true: Before simplification, the left-hand side of P Show that for each integer k ≥ (¹ - 1) (¹ - - - ) (¹ - 1) -- (¹ - ( and the right-hand side of P(k) is ,if P(K) is true, then P 2k [The inductive hypothesis is that the two sides of P(k) are equal.] We must show that P (¹ - - - ) (¹ - - - ) ( ¹ - - - - --- (¹ - 2k Hence, P 1 After substitution from the inductive hypothesis, the left-hand side of P ))·(¹-T ™)-(( Before simplification, the right-hand side of P When the left- and right-hand sides of P is true. In other words, we must show that the left and right-hand sides of P and the next-to-last factor in the left-hand side is (1-2). So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is (1 - - - ) (¹ - 3 ) (¹ - - ) --- (1 - 1) (¹ · 2k ])) ₁ (¹-²) 2k(k+ 1) is true, which completes the inductive step. 2(x + 1) is true: Let k be any integer with k z 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(K) is +1 and the right-hand side is U 2-2 becomes 1 are simplified, both can be shown to equal After simplification, both sides can be shown to equal are equal. The left-hand side of P . Thus, P() is true.
Use mathematical induction to prove the following statement. n+1 ≥ 2₁ (1-2) (₁-3) (1 - ²2 ) -^21 Proof (by mathematical induction): Let the property P(n) be the equation (¹-2)(¹-3)... - 2²2) = ² + 1. 20 We will show that P(n) is true for every integer n z. For every integer n 2 Show that P) is true: Before simplification, the left-hand side of P Show that for each integer k ≥ (¹ - 1) (¹ - - - ) (¹ - 1) -- (¹ - ( and the right-hand side of P(k) is ,if P(K) is true, then P 2k [The inductive hypothesis is that the two sides of P(k) are equal.] We must show that P (¹ - - - ) (¹ - - - ) ( ¹ - - - - --- (¹ - 2k Hence, P 1 After substitution from the inductive hypothesis, the left-hand side of P ))·(¹-T ™)-(( Before simplification, the right-hand side of P When the left- and right-hand sides of P is true. In other words, we must show that the left and right-hand sides of P and the next-to-last factor in the left-hand side is (1-2). So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is (1 - - - ) (¹ - 3 ) (¹ - - ) --- (1 - 1) (¹ · 2k ])) ₁ (¹-²) 2k(k+ 1) is true, which completes the inductive step. 2(x + 1) is true: Let k be any integer with k z 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(K) is +1 and the right-hand side is U 2-2 becomes 1 are simplified, both can be shown to equal After simplification, both sides can be shown to equal are equal. The left-hand side of P . Thus, P() is true.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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