Arrange the given steps in the correct order to prove that 3" 6, 3k 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" 6, 3k 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.
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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Please help me with these two questions. I am having trouble understanding what to do.
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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
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.
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