Adapt the formula 1 + 2 + 3 + ... + (n − 1) + n = nx(n+1) 2 a polynomial in terms of n and in terms of a triangular number. 1) S(n) = 4+5+6+...+n+ (n + 1). TO=1+2+3+ intc^tis-1-2-3 =1+2+3+1 = (n +D(~+2) 60 2 to find the following sums. Represent the answe ·le
Adapt the formula 1 + 2 + 3 + ... + (n − 1) + n = nx(n+1) 2 a polynomial in terms of n and in terms of a triangular number. 1) S(n) = 4+5+6+...+n+ (n + 1). TO=1+2+3+ intc^tis-1-2-3 =1+2+3+1 = (n +D(~+2) 60 2 to find the following sums. Represent the answe ·le
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
Is number 1 correct? If not can you help me with it?
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### Educational Mathematics Quiz
#### Course: MATH 641
**Take-Home Quiz 1.01b**
**Name:** Patrice Walker
**Points:** 12
**Total Score:** 5 / 12
---
**Directions:**
1. Show complete work to earn appropriate credit.
2. You may check your answers with a classmate.
3. Solve each problem in the space provided.
---
**Problem 1:**
Adapt the formula \(1 + 2 + 3 + \ldots + (n-1) + n = \frac{n \times (n+1)}{2}\) to find the following sums. Represent the answers as a polynomial in terms of \(n\) in terms of a triangular number.
**Solution:**
\[ S(n) = 4 + 5 + 6 + \ldots + n + (n+1) \]
Steps to solve:
\[
\begin{aligned}
&= (1 + 2 + 3 + \ldots + (n+1)) - (1 + 2 + 3) \\
&= \frac{(n+1)(n+2)}{2} - 6 \\
&= \frac{n^2 + 3n + 2}{2} - 6 \\
&= \frac{n^2 + 3n + 2 - 12}{2} \\
&= \frac{n^2 + 3n - 10}{2} \\
\end{aligned}
\]
Thus,
\[ S(n) = \frac{n^2 + 3n - 10}{2} \]
**Problem 2:**
\[ D(n) = (n-3) + (n-3) + (n-4) + (n-5) + \ldots + 3 + 2 + 1 \]
---
Note: Ensure to review the steps above, and check each calculation for accuracy. Collaboration with classmates is encouraged, but the representation of your individual understanding is essential.
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Transcribed Image Text:---
### Educational Mathematics Quiz
#### Course: MATH 641
**Take-Home Quiz 1.01b**
**Name:** Patrice Walker
**Points:** 12
**Total Score:** 5 / 12
---
**Directions:**
1. Show complete work to earn appropriate credit.
2. You may check your answers with a classmate.
3. Solve each problem in the space provided.
---
**Problem 1:**
Adapt the formula \(1 + 2 + 3 + \ldots + (n-1) + n = \frac{n \times (n+1)}{2}\) to find the following sums. Represent the answers as a polynomial in terms of \(n\) in terms of a triangular number.
**Solution:**
\[ S(n) = 4 + 5 + 6 + \ldots + n + (n+1) \]
Steps to solve:
\[
\begin{aligned}
&= (1 + 2 + 3 + \ldots + (n+1)) - (1 + 2 + 3) \\
&= \frac{(n+1)(n+2)}{2} - 6 \\
&= \frac{n^2 + 3n + 2}{2} - 6 \\
&= \frac{n^2 + 3n + 2 - 12}{2} \\
&= \frac{n^2 + 3n - 10}{2} \\
\end{aligned}
\]
Thus,
\[ S(n) = \frac{n^2 + 3n - 10}{2} \]
**Problem 2:**
\[ D(n) = (n-3) + (n-3) + (n-4) + (n-5) + \ldots + 3 + 2 + 1 \]
---
Note: Ensure to review the steps above, and check each calculation for accuracy. Collaboration with classmates is encouraged, but the representation of your individual understanding is essential.
---
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