Problem 113 - College Algebra

Question

Exercises will help you prepare for the material covered in the next section. In Exercises \(112-113,\) show that $$ 1+2+3+\cdots+n-\frac{n(n+1)}{2} $$is true for the given value of \(n .\) $$n-5 \text { : Show that } 1+2+3+4+5-\frac{5(5+1)}{2}.$$

Step-by-Step Solution

Answer

Yes, the equation, 1+2+3+...+5-\(\frac{5(5+1)}{2}\) is true as both sides of the equation equal to 15.

Step 1: Calculate the left side

The equation mentions a sum of numbers from 1 to \(n\). Specifying \(n = 5\) for this exercise, the left side of the equation becomes \(1 + 2 + 3 + 4 + 5 = 15\).

Step 2: Calculate the right side

The right side is given by \(\frac{n(n+1)}{2}\) where \(n = 5\). Thus, the result is \(\frac{5(5+1)}{2} = 15\).

Step 3: Compare the results

Comparing the results obtained from the left side (15) and the right side (15), it is seen that they are indeed equal. Thus, the original statement holds true for \(n = 5\).

Key Concepts

Arithmetic Series

An arithmetic series is a sequence of numbers where the difference between any two consecutive terms is constant. This difference is called the common difference. For example, in the series 1, 2, 3, 4, 5, the common difference is 1.

This type of sequence can be written in general form as: first term, second term plus the common difference, third term plus twice the common difference, and so on. In mathematical terms, if the first term is denoted by \(a_1\) and the common difference by \(d\), the nth term \(a_n\) can be expressed as \(a_1 + (n-1) \cdot d\).

If you wish to find the sum of an arithmetic series, you can use a specific formula derived from adding the terms in pairs, from opposite ends of the series. For example, for the series starting with 1 and ending with \(n\), the sum \(S_n\) is calculated as:

• \(S_n = \frac{n}{2} \cdot (a_1 + a_n)\)

This formula is incredibly useful because it allows you to calculate the sum of a long series without adding each individual term.

Sum of Integers

The sum of the first \(n\) positive integers can often be a pivotal part of arithmetic series studies. Given the natural series 1 + 2 + 3 + ... + \(n\), this sum is seen frequently in problems involving sequential addition.

An effective approach to find this sum involves using the formula:

• \(S_n = \frac{n(n+1)}{2}\)

To understand why this formula works, consider pairing terms from the beginning and end of the sequence: \((1 + n), (2 + (n-1)), (3 + (n-2))...\). Each of these pairs sums to \(n + 1\), and there are \(\frac{n}{2}\) such pairs if \(n\) is even, or \(\frac{n+1}{2}\) if \(n\) is odd.

This formula provides a quick solution, eliminating the need to perform repetitive calculations every time you encounter a sum of integers.

Algebraic Proof

An algebraic proof uses algebraic manipulations to establish the truth of a mathematical statement. It usually involves several steps that incrementally build the solution.

In a given exercise where it is shown that \(1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}\), the approach often involves:

• Calculating the sum on both sides individually.
• Comparing the results to verify equality.

In the example with \(n=5\):

• The left side is calculated to be \(1 + 2 + 3 + 4 + 5 = 15\).
• The right side, using the formula \(\frac{n(n+1)}{2}\), is \(\frac{5(5+1)}{2} = 15\).
• Since both sides equal 15, the equality holds true.

This process not only confirms the initial statement but showcases the utility of algebraic formulas in simplifying complex problems. Proofs are foundational in mathematics, providing assurances of consistency and accuracy.