Discrete Math **Title: Proving the Sum of Consecutive Natural Numbers****Description:**Explore how to prove the formula for the sum of the first $ n $ natural numbers using the well-ordered property.**Mathematical Statement:**Show, using the well ordered property, that the sum of the first $ n $ numbers can be expressed as:\[1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2}\]**Explanation:**In this problem, we aim to demonstrate the formula which calculates the sum of a sequence of consecutive numbers starting from 1 and ending at $ n $. We apply the well-ordered property, often used in proofs involving natural numbers, which states that every non-empty set of natural numbers has a least element. This property is essential for constructing mathematical demonstrations such as induction proofs.

Answered: Show, using the well ordered property,…

Math

Advanced Math

Show, using the well ordered property, that Give an 1+2+3+4++n= n(n+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

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Question

Discrete Math

**Title: Proving the Sum of Consecutive Natural Numbers**

**Description:**

Explore how to prove the formula for the sum of the first $ n $ natural numbers using the well-ordered property.

**Mathematical Statement:**

Show, using the well ordered property, that the sum of the first $ n $ numbers can be expressed as:

\[
1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2}
\]

**Explanation:**

In this problem, we aim to demonstrate the formula which calculates the sum of a sequence of consecutive numbers starting from 1 and ending at $ n $. We apply the well-ordered property, often used in proofs involving natural numbers, which states that every non-empty set of natural numbers has a least element. This property is essential for constructing mathematical demonstrations such as induction proofs.

Expert Solution

Step 1: Introduction of the given problem

Using well ordering property we have to show that $1 plus 2 plus 3 plus 4 plus.. plus n equals fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction$

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