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range the given steps in the correct order to prove that Rank the options below. k + = 1-2 k-(k+1) k+1 is true for all n>0 using the concept of mathematical induction. Assume that for some k> 0, 1212 1 1 + = 2-3 2-3 k-(k+1) k (k+1) k+ 蛋雞 By adding (k+1)(k+2) (k+1)(k+2) (k+1)(k+2) on both the sides, we get 12+. + = k-(k+1) k (k+1) (k+1)(k+2) (k+1)(k+2) (k+1)(k+2) 1 For n=1, the left-hand side of the theorem is 12 12 = and the right-hand side = +1 + = k(k+2)+1 (k+1)(k+2) Therefore, = = k²+2k+1 (k+1)(k+2) k+1 k+2 Therefore, we have + left hand-side = k k+1 k+ k(k+2)+1 (k+1)(k+2) k-(k+1) (k+1)(k+2) by the inductive hypothesis. k²+2k+1 k+1 k(k+2)+1 (k+1)(k+2) k+2 (k+1)(k+2) We have completed both the basis step and the inductive step; so, by the principle of mathematical induction, the statement is true for every positive integer n. + 1/2+233 + k-(k+1) (+1)+ +(k+1)(k+2) = k(k+2)+1 (k+1)(k+2) This is true because each term on

range the given steps in the correct order to prove that Rank the options below. k + = 1-2 k-(k+1) k+1 is true for all n>0 using the concept of mathematical induction. Assume that for some k> 0, 1212 1 1 + = 2-3 2-3 k-(k+1) k (k+1) k+ 蛋雞 By adding (k+1)(k+2) (k+1)(k+2) (k+1)(k+2) on both the sides, we get 12+. + = k-(k+1) k (k+1) (k+1)(k+2) (k+1)(k+2) (k+1)(k+2) 1 For n=1, the left-hand side of the theorem is 12 12 = and the right-hand side = +1 + = k(k+2)+1 (k+1)(k+2) Therefore, = = k²+2k+1 (k+1)(k+2) k+1 k+2 Therefore, we have + left hand-side = k k+1 k+ k(k+2)+1 (k+1)(k+2) k-(k+1) (k+1)(k+2) by the inductive hypothesis. k²+2k+1 k+1 k(k+2)+1 (k+1)(k+2) k+2 (k+1)(k+2) We have completed both the basis step and the inductive step; so, by the principle of mathematical induction, the statement is true for every positive integer n. + 1/2+233 + k-(k+1) (+1)+ +(k+1)(k+2) = k(k+2)+1 (k+1)(k+2) This is true because each term on

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Arrange the given steps in the correct order to prove that Rank the options below. + 1-2 2-3 k = is true for all n> O using the concept of mathematical induction. k-(k+1) k+1 Assume that for some k>0, 12 k k + + + = 2-3 2-3 k. (k+1) k (k+1) k+1 k+1 1 1 + + 2-3 2-3 k. (k+1) k (k+1) (k+1)(k+2) (k+1)(k+2) k+1 k+1 (k+1)(k+2) By adding (k+1)(k+2) (k+1) (k+2) on both the sides, we get 1212+213 1 (k+1)(k+2) 1 For n = 1, the left-hand side of the theorem is 12 12 n n = 22 and the right-hand side = n+1 n+1 12 12 Therefore, we have 2+3+...+ + k 1-2 k k-(k+1) (k+1)(k+2) k(k+2)+1 1 (k+1)(k+2) 1-2 k(k+2)+1 +23+...+ + This is true because each term on k-(k+1) (k+1)(k+2) (k+1)(k+2) left hand-side = +1 +1, by the inductive hypothesis. Therefore, k(k+2)+1 (k+1)(k+2) k²+2k+1 (k+1)(k+2) k+1 k(k+2)+1 k+2 (k+1)(k+2) k²+2k+1 (k+1)(k+2) k+1 k+2 We have completed both the basis step and the inductive step; so, by the principle of mathematical induction, the statement is true for every positive integer n.
Transcribed Image Text:Arrange the given steps in the correct order to prove that Rank the options below. + 1-2 2-3 k = is true for all n> O using the concept of mathematical induction. k-(k+1) k+1 Assume that for some k>0, 12 k k + + + = 2-3 2-3 k. (k+1) k (k+1) k+1 k+1 1 1 + + 2-3 2-3 k. (k+1) k (k+1) (k+1)(k+2) (k+1)(k+2) k+1 k+1 (k+1)(k+2) By adding (k+1)(k+2) (k+1) (k+2) on both the sides, we get 1212+213 1 (k+1)(k+2) 1 For n = 1, the left-hand side of the theorem is 12 12 n n = 22 and the right-hand side = n+1 n+1 12 12 Therefore, we have 2+3+...+ + k 1-2 k k-(k+1) (k+1)(k+2) k(k+2)+1 1 (k+1)(k+2) 1-2 k(k+2)+1 +23+...+ + This is true because each term on k-(k+1) (k+1)(k+2) (k+1)(k+2) left hand-side = +1 +1, by the inductive hypothesis. Therefore, k(k+2)+1 (k+1)(k+2) k²+2k+1 (k+1)(k+2) k+1 k(k+2)+1 k+2 (k+1)(k+2) k²+2k+1 (k+1)(k+2) k+1 k+2 We have completed both the basis step and the inductive step; so, by the principle of mathematical induction, the statement is true for every positive integer n.
Arrange the given steps in the correct order to prove that 3" <n! if n is an integer greater than 6, using mathematical induction. Rank the options below. Suppose that for some k> 6, 3k <k! 3k <k! 3+1=3.3k 3+1 = 3.3k 3+1 < (k+1) k! 3*+1 < (k+1). k! For n=7, 37=2187 < 7! = 5040 37 = 2187<7! = 5040. 3+1 (k+1)! 3+1 < (k+1)! 3+(k+1)-3 3+1 < (k+1).3k
Transcribed Image Text:Arrange the given steps in the correct order to prove that 3" <n! if n is an integer greater than 6, using mathematical induction. Rank the options below. Suppose that for some k> 6, 3k <k! 3k <k! 3+1=3.3k 3+1 = 3.3k 3+1 < (k+1) k! 3*+1 < (k+1). k! For n=7, 37=2187 < 7! = 5040 37 = 2187<7! = 5040. 3+1 (k+1)! 3+1 < (k+1)! 3+(k+1)-3 3+1 < (k+1).3k
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