Arrange the given steps in the correct order to prove that +23+++1)=+1 is true for all n> 0 using the concept c Rank the options below. 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. By adding (+1+2) (+1)(+2) on both the sides, we get 1/2 √2+23 23+-+=(√²+1) = (²+1) + (x+1)(x+2)(x+1)(x+2) = +1 +1 + (+1)(x+2) (x+1)(x+2) For n=1, the left-hand side of the theorem is and the right-hand side of mathematical induction. W n+1 Assume that for some k> 0, 1/2 1/2 + 23 23 + -+ 2+1) +++ o Therefore, we have 2 +2²3 + ... + . √²+1)+(²+1)(2+2) = (2+1+2) 1/2 +23+...+(²+1) + (+1)(x+2) = (2+1+2). This is true because each term on left hand-side by the inductive hypothesis. (k+1)(+2) (k+1)(+2) = 4+2 (4+1)(4+2) = (2+1)(4+2) = 4+2 k-(k+1) k²+2k+1 k+1 A(+2)+1 A(+2)+1 Therefore k+1
Arrange the given steps in the correct order to prove that +23+++1)=+1 is true for all n> 0 using the concept c Rank the options below. 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. By adding (+1+2) (+1)(+2) on both the sides, we get 1/2 √2+23 23+-+=(√²+1) = (²+1) + (x+1)(x+2)(x+1)(x+2) = +1 +1 + (+1)(x+2) (x+1)(x+2) For n=1, the left-hand side of the theorem is and the right-hand side of mathematical induction. W n+1 Assume that for some k> 0, 1/2 1/2 + 23 23 + -+ 2+1) +++ o Therefore, we have 2 +2²3 + ... + . √²+1)+(²+1)(2+2) = (2+1+2) 1/2 +23+...+(²+1) + (+1)(x+2) = (2+1+2). This is true because each term on left hand-side by the inductive hypothesis. (k+1)(+2) (k+1)(+2) = 4+2 (4+1)(4+2) = (2+1)(4+2) = 4+2 k-(k+1) k²+2k+1 k+1 A(+2)+1 A(+2)+1 Therefore k+1
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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Transcribed Image Text:Arrange the given steps in the correct order to prove that 12 + 2-3
+ ... +
Rank the options below.
By adding (k+1)(k+2) (k+1)(k+2) on both the sides, we get
=
For n = 1, the left-hand side of the theorem is 1-2 1-2
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
Therefore, we have + + ... +
1.2 2.3
k(k+2)+1
k²+2k+1
(k+1)(k+2) (k+1)(k+2) k+2
Therefore,
1
+
k.(k+1) (k+1)(k+2)
k+1 k(k+2)+1
(k+1)(k+2)
1
+
Assume that for some k> 0, 1-2 1-2
k. (k+1)
1 1
+ ... +
2.3 2.3
k(k+ 2) + 1
(k+ 1)(k+2)
k²+2k+1
(k+1)(k+2)
k
k
+
+ +
1.2 1.2 2.3 2:3 k² (k+1) k² (k+1) + (k+1)(k+2) (k+1)(k+2) =k+1 k+1 + (k+1)(k+2) (k+1)(k+2)
=
and the right-hand side =
1
1
k. (k+1) k. (k+1)
k
is true for all n> 0 using the concept of mathematical induction.
k+1
k+1
k+2
=
1 1/2 +
n
n
n+1 n+1
+ ... +
2.3
k
k
k+1 k+1
k.(k+1)
=
+
1
(k+1)(k+2)
k(k+ 2) + 1
(k+ 1)(k+2)
. This is true because each term on left hand-side =
k
k+1
k
k+1
by the inductive hypothesis.
,
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