Question

Sum of integers Let \(S(n)=1+2+\ldots+n,\) where \(n\) is a positive integer. It can be shown that \(S(n)=n(n+1) / 2\) a. Make a table of \(S(n),\) for \(n=1,2, \ldots, 10\) b. How would you describe the domain of this function? c. What is the least value of \(n\) for which \(S(n)>1000 ?\)

Short answer

Answer: The least value of n for which the sum of the first n positive integers exceeds 1000 is 45.

Step-by-step solution

(a. Create a table of S(n) for n = 1 to 10)

To create the table of S(n) for n = 1 to 10, let's substitute the values of n from 1 to 10 in the formula \(S(n) = \frac{n(n + 1)}{2}\): For \(n=1\): \(S(1) = \frac{1(1 + 1)}{2} = 1\) For \(n=2\): \(S(2) = \frac{2(2 + 1)}{2} = 3\) For \(n=3\): \(S(3) = \frac{3(3 + 1)}{2} = 6\) Repeat this process for n = 4 to 10 to complete the table.

(b. Describe the domain of the function S(n))

The domain of a function consists of all the input values for which the function is defined. In our case, the function S(n) gives the sum of the first n positive integers. Considering that n represents the positive integers, the domain of the function S(n) is the set of all positive integers. Mathematically, this can be written as \(D = \{n \in \mathbb{Z^+}\}\), where \(\mathbb{Z^+}\) represents the set of positive integers.

(c. Find the least value of n for which S(n) > 1000)

To find the smallest integer n for which \(S(n) > 1000\), we need to solve the inequality \(\frac{n(n + 1)}{2} > 1000\). We can start by multiplying both sides by 2 to get rid of the fraction: \(n(n + 1) > 2000\) Now, we can start checking the integer values of n, starting with the smallest possible value: For \(n = 1\), we have \(1 (1+1) = 2 < 2000\) Keep increasing the value of n and checking if the inequality holds: ... For \(n = 44\), we have \(44 (44+1) = 1980 < 2000\) For \(n = 45\), we have \(45 (45+1) = 2070 > 2000\) Since the inequality holds true for \(n = 45\), the least value of n for which \(S(n) > 1000\) is \(n = 45\).

Key concepts

Arithmetic Series

An arithmetic series is the summation of terms in an arithmetic sequence, where each term is generated by adding a constant value to the previous term. In the context of summing integers from 1 to a given positive integer n, the sequence is 1, 2, 3, ..., n, where the common difference is simply 1. The series can be expressed as the sum, written compactly as:
\( S(n) = 1 + 2 + 3 + ... + n \).
This series is vital in numerous mathematical applications and is particularly elegant because the formula to calculate its sum is straightforward:
\( S(n) = \frac{n(n + 1)}{2} \).
This formula is derived from pairing terms equidistant from the start and end of the series (e.g., 1 + n, 2 + (n-1), etc.), with each pair summing to \( (n + 1) \) and having a total of \( \frac{n}{2} \) such pairs.

Function Domain

The domain of a function refers to the complete set of 'input' values for which the function is defined. When considering functions like an arithmetic series, the domain is especially crucial. For the function
\( S(n) = \frac{n(n + 1)}{2} \),
which gives the sum of the first n positive integers, the domain consists of all positive integers since you cannot sum a negative, zero, or fractional quantity of integers in this context. Represented in mathematical notation, it becomes
\( D = \{n \: n \: \in \: \mathbb{Z^+}\} \),
where \( \mathbb{Z^+} \) denotes the set of all positive integers. Understanding the domain is crucial for correctly applying the function and interpreting its results.

Inequality Solving

Solving inequalities is a fundamental aspect of algebra that involves finding the values of variables that make an inequality statement true. The process is similar to solving equations, but with extra attention to the direction of the inequality when multiplying or dividing by negative numbers.
In the exercise at hand, we are tasked with finding the smallest integer n such that \( S(n) > 1000 \), which translates into an inequality:
\( \frac{n(n + 1)}{2} > 1000 \).
To solve it, one approach is to manipulate the inequality algebraically to isolate n. Alternatively, we might estimate and verify by incrementing n and checking the condition, a more intuitive, albeit less efficient, method. As we increment n, we seek the threshold where the inequality becomes true. Once found, we establish the smallest integer needed for the sum of the first n integers to exceed 1000, emphasizing the importance of inequality solving in identifying numeric boundaries within a given context.