6. Use mathematical induction to prove that 2+5+8+11+.....+(3n-1)=n(3n+1)/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
Question
Discrete Mathematics: Please Help me with Question 6 (See attachment)
![## Mathematical Problems and Solutions
### 1. Compute the Following:
a) \(\prod_{n=2}^{58} (-1)^n\)
b) \(2 + 5 + 8 + 11 + \ldots + 302 + 305\)
c) \(\sum_{i=0}^{100} \frac{i}{1000!}\)
d) \(\frac{1000!}{998!}\)
### 2. Recursive Function Problem:
For the following recursive functions, find \(f(1)\), \(f(2)\), \(f(3)\), and \(f(4)\)
- Given: \(f(0) = 2\)
\[f(k) = k - (f(k-1))^2\]
- Calculations:
- \(F(1) = 1 - (f(0))^2 = 1 - 4 = -3\)
- \(F(2) = 2 - (f(1))^2 = 2 - (-3)^2 = -7\)
- \(F(3) = 3 - (f(2))^2 = 3 - 49 = -46\)
- \(F(4) = 4 - (f(3))^2 = 4 - (−46)^2 = 4 - 2116 = -2112\)
### 3. Prove Relation:
Prove that \(S_n = \frac{n^2 + n}{2}\) is a solution of the recursive relation
- \(S_1 = 1\)
- \(S_k = S_{k-1} + k \quad \text{for } k > 1 \)
### 4. Solve the Recursive Relation Using Pattern Recognition:
Given:
- \(f(0) = 4\)
\[f(k) = 5 + 1.1f(k-1) \quad \text{for } k > 0\]
a) Evaluate \(f(10)\)
### 5. Divisibility Problem:
Prove: \(n^3 + 5n\) is divisible by 6 for all integer \(n \geq 0\).
### 6. Mathematical Induction:
Prove by induction:
\[2 + 5 + 8 + 11 + \ldots + (](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd6805e19-16c9-4240-8da2-d1d02f6f503a%2F4cea5ffa-c70b-4c54-ae0f-6c67391e6c1b%2Fmprdux4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Mathematical Problems and Solutions
### 1. Compute the Following:
a) \(\prod_{n=2}^{58} (-1)^n\)
b) \(2 + 5 + 8 + 11 + \ldots + 302 + 305\)
c) \(\sum_{i=0}^{100} \frac{i}{1000!}\)
d) \(\frac{1000!}{998!}\)
### 2. Recursive Function Problem:
For the following recursive functions, find \(f(1)\), \(f(2)\), \(f(3)\), and \(f(4)\)
- Given: \(f(0) = 2\)
\[f(k) = k - (f(k-1))^2\]
- Calculations:
- \(F(1) = 1 - (f(0))^2 = 1 - 4 = -3\)
- \(F(2) = 2 - (f(1))^2 = 2 - (-3)^2 = -7\)
- \(F(3) = 3 - (f(2))^2 = 3 - 49 = -46\)
- \(F(4) = 4 - (f(3))^2 = 4 - (−46)^2 = 4 - 2116 = -2112\)
### 3. Prove Relation:
Prove that \(S_n = \frac{n^2 + n}{2}\) is a solution of the recursive relation
- \(S_1 = 1\)
- \(S_k = S_{k-1} + k \quad \text{for } k > 1 \)
### 4. Solve the Recursive Relation Using Pattern Recognition:
Given:
- \(f(0) = 4\)
\[f(k) = 5 + 1.1f(k-1) \quad \text{for } k > 0\]
a) Evaluate \(f(10)\)
### 5. Divisibility Problem:
Prove: \(n^3 + 5n\) is divisible by 6 for all integer \(n \geq 0\).
### 6. Mathematical Induction:
Prove by induction:
\[2 + 5 + 8 + 11 + \ldots + (
Expert Solution

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 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,

