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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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