Use mathematical induction to prove the following statement. n+1 : 2, (¹ - 2 ) (¹ - ¹3 ) --- (¹ - 22 ) -^2¹ Proof (by mathematical induction): Let the property P(n) be the equation (₁-21) (¹ - 1) (¹ - 1) - We will show that P(n) is true for every integer n 2 For every integer n 2 2, Show that P Show that for each integer k is true: Before simplification, the left-hand side of Pis (¹-2)(¹-3) (¹ - 1) and the right-hand side of P(k) is We must show that |- (¹ - 2 ) ) ( ¹ - 3 - ) ( ¹ - - - - - - - (¹ - 1- n+1 2n if P(k) is true, then P 2k [The inductive hypothesis is that the two sides of P(k) are equal.] PC When the left- and right-hand sides of P 1 +1 Before simplification, the right-hand side of P and the next-to-last factor in the left-hand side is (1-21)- (¹ - 2 ) (¹ - ² ) (¹ - - ) --- (1 - 2 ) ( ¹ - ( 1 is true. In other words, we must show that the left and right-hand sides of P After substitution from the inductive hypothesis, the left-hand side of P I).(+-+ T)-( 1 2k -O). 2k(k+1) 1- 2(k+1) So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is and the right-hand side is becomes 1 +1. After simplification, both sides can be shown to equal 2.2 are simplified, both can be shown to equal are equal. The left-hand side of . Thus, P is true.
Use mathematical induction to prove the following statement. n+1 : 2, (¹ - 2 ) (¹ - ¹3 ) --- (¹ - 22 ) -^2¹ Proof (by mathematical induction): Let the property P(n) be the equation (₁-21) (¹ - 1) (¹ - 1) - We will show that P(n) is true for every integer n 2 For every integer n 2 2, Show that P Show that for each integer k is true: Before simplification, the left-hand side of Pis (¹-2)(¹-3) (¹ - 1) and the right-hand side of P(k) is We must show that |- (¹ - 2 ) ) ( ¹ - 3 - ) ( ¹ - - - - - - - (¹ - 1- n+1 2n if P(k) is true, then P 2k [The inductive hypothesis is that the two sides of P(k) are equal.] PC When the left- and right-hand sides of P 1 +1 Before simplification, the right-hand side of P and the next-to-last factor in the left-hand side is (1-21)- (¹ - 2 ) (¹ - ² ) (¹ - - ) --- (1 - 2 ) ( ¹ - ( 1 is true. In other words, we must show that the left and right-hand sides of P After substitution from the inductive hypothesis, the left-hand side of P I).(+-+ T)-( 1 2k -O). 2k(k+1) 1- 2(k+1) So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is and the right-hand side is becomes 1 +1. After simplification, both sides can be shown to equal 2.2 are simplified, both can be shown to equal are equal. The left-hand side of . 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
Related questions
Question
need help
![Use mathematical induction to prove the following statement.
1
n+
22, (1 - 12/2) (¹ - 1) (¹ - 22 ) - 02/1
1
1
=
2n
Proof (by mathematical induction): Let the property P(n) be the equation
1
(¹₁ - 1212) (¹ - 3 2 ) - ( ¹₁ - 12/1 ) -
1
1
1
3²
We will show that P(n) is true for every integer n 2
Show that P
For every integer n ≥ 2, 1-
Show that for each integer k ≥
(¹ - 2 ) ( ¹ - - - ) ( ¹ - - - ) . · ( ¹ -
1
1
3²
and the right-hand side of P(k) is
is true: Before simplification, the left-hand side of P
We must show that P
1
(¹ - 1) (¹ - ¹) (¹ - 4- ) --- (¹
1
1
1 -
22
2k
, if P(k) is true, then P
[The inductive hypothesis is that the two sides of P(k) are equal.]
n+ 1
2n
2k
Hence, P
1
When the left and right-hand sides of P
+ 1
and the next-to-last factor in the left-hand side is 1-
is (1-1/2)
1
(¹ - 2 ) ( ¹ - 3 - ) ( ¹ - - - - - - - (1 - 2 ) ( ¹
1-
1
1
121) (₁ - (
1-
2²
4²
k²
Before simplification, the right-hand side of P
1
is true. In other words, we must show that the left- and right-hand sides of P
After substitution from the inductive hypothesis, the left-hand side of P
1
I)·(²-²)³²) - (
1
2k
2k(k+1)
Y)
is
So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is
is
0(1-11
2k
is true, which completes the inductive step.
2 (K + 1)
+1
and the right-hand side is
is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is
becomes
1
are simplified, both can be shown to equal
+ 1
2.2
After simplification, both sides can be shown to equal
are equal. The left-hand side of P
is
Thus, P
is true.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3cf874b-7a7b-478f-a7b8-421442e72224%2F38294ce9-3528-47f3-8338-f788dd9ac185%2F6sgryo_processed.png&w=3840&q=75)
Transcribed Image Text:Use mathematical induction to prove the following statement.
1
n+
22, (1 - 12/2) (¹ - 1) (¹ - 22 ) - 02/1
1
1
=
2n
Proof (by mathematical induction): Let the property P(n) be the equation
1
(¹₁ - 1212) (¹ - 3 2 ) - ( ¹₁ - 12/1 ) -
1
1
1
3²
We will show that P(n) is true for every integer n 2
Show that P
For every integer n ≥ 2, 1-
Show that for each integer k ≥
(¹ - 2 ) ( ¹ - - - ) ( ¹ - - - ) . · ( ¹ -
1
1
3²
and the right-hand side of P(k) is
is true: Before simplification, the left-hand side of P
We must show that P
1
(¹ - 1) (¹ - ¹) (¹ - 4- ) --- (¹
1
1
1 -
22
2k
, if P(k) is true, then P
[The inductive hypothesis is that the two sides of P(k) are equal.]
n+ 1
2n
2k
Hence, P
1
When the left and right-hand sides of P
+ 1
and the next-to-last factor in the left-hand side is 1-
is (1-1/2)
1
(¹ - 2 ) ( ¹ - 3 - ) ( ¹ - - - - - - - (1 - 2 ) ( ¹
1-
1
1
121) (₁ - (
1-
2²
4²
k²
Before simplification, the right-hand side of P
1
is true. In other words, we must show that the left- and right-hand sides of P
After substitution from the inductive hypothesis, the left-hand side of P
1
I)·(²-²)³²) - (
1
2k
2k(k+1)
Y)
is
So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is
is
0(1-11
2k
is true, which completes the inductive step.
2 (K + 1)
+1
and the right-hand side is
is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is
becomes
1
are simplified, both can be shown to equal
+ 1
2.2
After simplification, both sides can be shown to equal
are equal. The left-hand side of P
is
Thus, P
is true.
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 7 steps with 6 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)
Recommended textbooks for you
![Advanced Engineering Mathematics](https://www.bartleby.com/isbn_cover_images/9780470458365/9780470458365_smallCoverImage.gif)
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
![Numerical Methods for Engineers](https://www.bartleby.com/isbn_cover_images/9780073397924/9780073397924_smallCoverImage.gif)
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
![Introductory Mathematics for Engineering Applicat…](https://www.bartleby.com/isbn_cover_images/9781118141809/9781118141809_smallCoverImage.gif)
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
![Advanced Engineering Mathematics](https://www.bartleby.com/isbn_cover_images/9780470458365/9780470458365_smallCoverImage.gif)
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
![Numerical Methods for Engineers](https://www.bartleby.com/isbn_cover_images/9780073397924/9780073397924_smallCoverImage.gif)
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
![Introductory Mathematics for Engineering Applicat…](https://www.bartleby.com/isbn_cover_images/9781118141809/9781118141809_smallCoverImage.gif)
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
![Mathematics For Machine Technology](https://www.bartleby.com/isbn_cover_images/9781337798310/9781337798310_smallCoverImage.jpg)
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
![Basic Technical Mathematics](https://www.bartleby.com/isbn_cover_images/9780134437705/9780134437705_smallCoverImage.gif)
![Topology](https://www.bartleby.com/isbn_cover_images/9780134689517/9780134689517_smallCoverImage.gif)