Question

Let \(T(n)=1^{2}+2^{2}+\cdots+n^{2}\) where \(n\) is a positive integer. It can be shown that \(T(n)=n(n+1)(2 n+1) / 6\) a. Make a table of \(T(n),\) 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 \(T(n)>1000 ?\)

Short answer

Question: Calculate the given sum of squares expression for values \(1\) to \(10\), describe the domain of the function, and find the least value of \(n\) for which the sum is greater than 1000. Answer: The sum of squares expression is calculated using the formula \(T(n)=\frac{n(n+1)(2n+1)}{6}\). For values \(n=1\) to \(10\), the sums are: \(1, 5, 14, 30, 55, 91, 140, 204, 285, 385\). The domain of the function is the set of all natural numbers (\(n \in \mathbb{N}\)). The least value of \(n\) for which \(T(n)>1000\) is \(n=14\).

Step-by-step solution

(Part a: Creating a table for \(T(n)\) for \(n=1\) to \(n=10\))

To create the table, calculate the values of \(T(n)\) using the given formula: \(T(n)=\frac{n(n+1)(2n+1)}{6}\). The table looks like this: | \(n\) | \(T(n)\) | |-----|---------------------| | 1 | \(\frac{1(1+1)(2+1)}{6} = 1\) | | 2 | \(\frac{2(2+1)(4+1)}{6} = 5\) | | 3 | \(\frac{3(3+1)(6+1)}{6} = 14\) | | 4 | \(\frac{4(4+1)(8+1)}{6} = 30\) | | 5 | \(\frac{5(5+1)(10+1)}{6} = 55\) | | 6 | \(\frac{6(6+1)(12+1)}{6} = 91\) | | 7 | \(\frac{7(7+1)(14+1)}{6} = 140\)| | 8 | \(\frac{8(8+1)(16+1)}{6} = 204\)| | 9 | \(\frac{9(9+1)(18+1)}{6} = 285\)| | 10 | \(\frac{10(10+1)(20+1)}{6} = 385\)| The table shows the values of \(T(n)\) for \(n=1\) to \(n=10\).

(Part b: Describing the domain of the function)

The domain of a function is a set of all possible input values (independent variable) for which the function is defined. In this case, the input value is \(n\). Since \(n\) is a positive integer (as specified in the problem statement), the domain of the function \(T(n)\) can be described as the set of all natural numbers, or mathematically, \(n \in \mathbb{N}\).

(Part c: Finding the least value of \(n\) for which \(T(n)>1000\))

To find the least value of \(n\) such that \(T(n)>1000\), we'll use the given formula for \(T(n)\) and solve for \(n\): \(T(n)=\frac{n(n+1)(2n+1)}{6}>1000\) We need to find the smallest integer \(n\) for which this inequality holds true. By trying out different values of \(n\) and plugging them into the inequality (similar to how we calculated the values for the table in part a), we can find the least value of \(n\): For \(n=20:\) \(T(20)=\frac{20(20+1)(40+1)}{6} = 2870\) For \(n=19:\) \(T(19)=\frac{19(19+1)(38+1)}{6} = 2470\) For \(n=18:\) \(T(18)=\frac{18(18+1)(36+1)}{6} = 2109\) For \(n=17:\) \(T(17)=\frac{17(17+1)(34+1)}{6} = 1785\) For \(n=16:\) \(T(16)=\frac{16(16+1)(32+1)}{6} = 1496\) For \(n=15:\) \(T(15)=\frac{15(15+1)(30+1)}{6} = 1240\) For \(n=14:\) \(T(14)=\frac{14(14+1)(28+1)}{6} = 1015\) For \(n=13:\) \(T(13)=\frac{13(13+1)(26+1)}{6} = 819\) Since \(T(13) = 819 < 1000\) and \(T(14) = 1015 > 1000\), we can conclude that the least value of \(n\) such that \(T(n)>1000\) is \(n=14\).

Key concepts

Positive Integers: Understanding Their Nature

Positive integers are whole numbers greater than zero. They are the basis of arithmetic and form the set \( \{ 1, 2, 3, \ldots \} \). Positive integers are involved in many mathematical problems and applications, often serving as simple and clear inputs for functions and calculations. In the context of summation of squares, positive integers are used as the values of \( n \) for which the sum of squares is calculated. For example, in the formula for \( T(n) \), where \( T(n) = 1^2 + 2^2 + \cdots + n^2 \), only positive integer values of \( n \) make sense because we're counting squares starting from 1. This foundational concept ensures that the function is defined and yields valid outputs. Positive integers simplify the process and are crucial for maintaining mathematical order and logic.

Arithmetic Series and Their Relevance

An arithmetic series is a sum of numbers with a constant difference between consecutive elements. However, in our case, we deal with a series of squared positive integers, which isn't strictly an arithmetic series but has its own unique characteristics.A major aspect of series like \( T(n) = 1^2 + 2^2 + \cdots + n^2 \) is the formula used to simplify the calculation. The function for \( T(n) \) utilizes the arithmetic properties of numbers augmented by their squared form, to efficiently compute the summation:

• For any positive integer \( n \), \( T(n) = \frac{n(n+1)(2n+1)}{6} \)
• This formula transforms a potentially long computation process into a straightforward calculation.

By recognizing patterns in these sums, mathematicians have derived formulas like \( T(n) \) to deal with very specific types of series operations, allowing us to bypass repetitive and extensive additions.

Inequality Analysis: Determining Boundaries

Inequality analysis helps us define conditions under which a certain result or property holds true. In solving the problem of finding the least \( n \) for which \( T(n) > 1000 \), we employ inequality analysis to systematically narrow down the potential values of \( n \) that satisfy this condition.To solve the inequality \( \frac{n(n+1)(2n+1)}{6} > 1000 \), we begin by testing values of \( n \) starting with an educated guess, usually based on prior analysis or calculations. A structured approach such as:

• Calculating \( T(n) \) for increasing values of \( n \)
• Identifying when the inequality finally holds true\( \( T(n) > 1000 \) \)

helps in pinpointing the exact boundary point. In this exercise, through computation, it was established that \( n = 14 \) is the smallest value where \( T(n) \) crosses the 1000 threshold. This technique is crucial in various fields such as optimization and analysis for determining precise operational limits.