Prove the following statement by mathematical induction. For every integer n2 0, 2" < (n + 2)!. Proof (by mathematical induction): Let P(n) be the inequality 2" < (n + 2)!. We will show that P(n) is true for every integer n 2 0. Show that P(0) is true: Before simplifying, the left-hand side of P(0) is 2 and the right-hand side is The fact that the statement is true can be deduced from that fact that 2° = 1. Show that for each integer k2 o, if P(k) is true, then P(k + 1) is true: Let k be any integer with k z 0, and suppose that P(k) is true. In other words, suppose that 2* < (*+ 2)! [This is P(k), the inductive hypothesis.] We must show that P(k + 1) is true. P(k + 1) is the inequality < (k+ 3)! Information about P(k + 1) can be deduced from the following steps. Identify the reason for each step. 2* < (k + 2)! 2*.2 < (k + 2)! - 2 2* +1 < (k + 2)! - 2 2* +1 < (k + 2)! - (k + 3) 2* +1 < (k + 3)! by the induction hypothesis v by basic algebra by basic algebra because 2
Prove the following statement by mathematical induction. For every integer n2 0, 2" < (n + 2)!. Proof (by mathematical induction): Let P(n) be the inequality 2" < (n + 2)!. We will show that P(n) is true for every integer n 2 0. Show that P(0) is true: Before simplifying, the left-hand side of P(0) is 2 and the right-hand side is The fact that the statement is true can be deduced from that fact that 2° = 1. Show that for each integer k2 o, if P(k) is true, then P(k + 1) is true: Let k be any integer with k z 0, and suppose that P(k) is true. In other words, suppose that 2* < (*+ 2)! [This is P(k), the inductive hypothesis.] We must show that P(k + 1) is true. P(k + 1) is the inequality < (k+ 3)! Information about P(k + 1) can be deduced from the following steps. Identify the reason for each step. 2* < (k + 2)! 2*.2 < (k + 2)! - 2 2* +1 < (k + 2)! - 2 2* +1 < (k + 2)! - (k + 3) 2* +1 < (k + 3)! by the induction hypothesis v by basic algebra by basic algebra because 2
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
Concept explainers
Quadratic Equation
When it comes to the concept of polynomial equations, quadratic equations can be said to be a special case. What does solving a quadratic equation mean? We will understand the quadratics and their types once we are familiar with the polynomial equations and their types.
Demand and Supply Function
The concept of demand and supply is important for various factors. One of them is studying and evaluating the condition of an economy within a given period of time. The analysis or evaluation of the demand side factors are important for the suppliers to understand the consumer behavior. The evaluation of supply side factors is important for the consumers in order to understand that what kind of combination of goods or what kind of goods and services he or she should consume in order to maximize his utility and minimize the cost. Therefore, in microeconomics both of these concepts are extremely important in order to have an idea that what exactly is going on in the economy.
Question
Please what is the answer of the left-hand side of P(0) is and the right-hand side is? And please show it clearly
![Prove the following statement by mathematical induction.
For every integer \( n \geq 0 \), \( 2^n < (n + 2)! \).
**Proof (by mathematical induction):** Let \( P(n) \) be the inequality \( 2^n < (n + 2)! \). We will show that \( P(n) \) is true for every integer \( n \geq 0 \).
**Show that \( P(0) \) is true:** Before simplifying, the left-hand side of \( P(0) \) is \( 2 \) and the right-hand side is \( 6 \). The fact that the statement is true can be deduced from the fact that \( 2^0 = 1 \).
**Show that for each integer \( k \geq 0 \), if \( P(k) \) is true, then \( P(k + 1) \) is true:** Let \( k \) be any integer with \( k \geq 0 \), and suppose that \( P(k) \) is true. In other words, suppose that:
\[ 2^k < (k + 2)! \]
[This is \( P(k) \), the inductive hypothesis.]
We must show that \( P(k + 1) \) is true. \( P(k + 1) \) is the inequality:
\[ 2^{k+1} < (k + 3)! \]
Information about \( P(k + 1) \) can be deduced from the following steps. Identify the reason for each step.
1. \( 2^k < (k + 2)! \) (by the induction hypothesis)
2. \( 2 \cdot 2^k < 2 \cdot (k + 2)! \) (by basic algebra)
3. \( 2^{k+1} < 2 \cdot (k + 2)! \) (by basic algebra)
4. \( 2^{k+1} < (k + 2 + k + 3) \) (because 2 = k + 3)
5. \( 2^{k+1} < (k + 3)! \) (by basic algebra)
Since \( 2^{k+1} < (k + 3)! \), the inequality for \( P](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc357075-d330-4116-93bc-b2819b26405c%2F10cd96f5-c1d6-4f7a-ab15-b15afb6a2e83%2F1gcvdo_processed.png&w=3840&q=75)
Transcribed Image Text:Prove the following statement by mathematical induction.
For every integer \( n \geq 0 \), \( 2^n < (n + 2)! \).
**Proof (by mathematical induction):** Let \( P(n) \) be the inequality \( 2^n < (n + 2)! \). We will show that \( P(n) \) is true for every integer \( n \geq 0 \).
**Show that \( P(0) \) is true:** Before simplifying, the left-hand side of \( P(0) \) is \( 2 \) and the right-hand side is \( 6 \). The fact that the statement is true can be deduced from the fact that \( 2^0 = 1 \).
**Show that for each integer \( k \geq 0 \), if \( P(k) \) is true, then \( P(k + 1) \) is true:** Let \( k \) be any integer with \( k \geq 0 \), and suppose that \( P(k) \) is true. In other words, suppose that:
\[ 2^k < (k + 2)! \]
[This is \( P(k) \), the inductive hypothesis.]
We must show that \( P(k + 1) \) is true. \( P(k + 1) \) is the inequality:
\[ 2^{k+1} < (k + 3)! \]
Information about \( P(k + 1) \) can be deduced from the following steps. Identify the reason for each step.
1. \( 2^k < (k + 2)! \) (by the induction hypothesis)
2. \( 2 \cdot 2^k < 2 \cdot (k + 2)! \) (by basic algebra)
3. \( 2^{k+1} < 2 \cdot (k + 2)! \) (by basic algebra)
4. \( 2^{k+1} < (k + 2 + k + 3) \) (because 2 = k + 3)
5. \( 2^{k+1} < (k + 3)! \) (by basic algebra)
Since \( 2^{k+1} < (k + 3)! \), the inequality for \( P
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
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…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
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…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
