Problem 7 - Essential Calculus

Question

Write the first five terms of the sequence. \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\)

Step-by-Step Solution

Answer

The first five terms of the sequence are -1, -1/4, -1/9, -1/16, -1/25

Step 1: Substitute n=1

Put \(n=1\) into \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\), we have \(a_{1}=\frac{(-1)^{1(1+1) / 2}}{1^{2}}=\frac{(-1)^{1}}{1}=-1\)

Step 2: Substitute n=2

Put \(n=2\) into \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\), we have \(a_{2}=\frac{(-1)^{2(2+1) / 2}}{2^{2}}=\frac{(-1)^{3}}{4}= -\frac{1}{4}\)

Step 3: Substitute n=3

Put \(n=3\) into \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\), we have \(a_{3}=\frac{(-1)^{3(3+1) / 2}}{3^{2}}=\frac{(-1)^{3}}{9}=-\frac{1}{9}\)

Step 4: Substitute n=4

Put \(n=4\) into \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\), we have \(a_{4}=\frac{(-1)^{4(4+1) / 2}}{4^{2}}=\frac{(-1)^{5}}{16}= -\frac{1}{16}\)

Step 5: Substitute n=5

Put \(n=5\) into \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\), we have \(a_{5}=\frac{(-1)^{5(5+1) / 2}}{5^{2}}=\frac{(-1)^{5}}{25}= -\frac{1}{25}\)

Key Concepts

Sequence Formula

A sequence in mathematics is an ordered list of numbers that follows a specific rule or pattern. Each number in the list is called a term. The sequence formula helps us determine the general term, usually denoted by \(a_n\), which can calculate any term in the sequence by substituting different values of \(n\).

For the given exercise, the sequence formula is:

• \(a_n = \frac{(-1)^{n(n+1)/2}}{n^2}\)

This complex formula involves several components:

• The exponent \((-1)^{n(n+1)/2}\) creates alternating positive and negative signs. It controls the sign of each term.
• The denominator \(n^2\) determines the magnitude of the term by using the square of the sequence position \(n\).

This combination of sign and magnitude gives us a series of fractions with alternating signs, as seen in the exercise's step-by-step solution.

Alternating Series

An alternating series is a mathematical series where the terms alternate in sign. This is clearly presented in the given sequence where the sign is determined by \((-1)^{n(n+1)/2}\). Alternating series are important in mathematics as they often appear in calculus and analysis, especially in infinite series and convergence tests.

In this exercise, each term of the sequence is produced with an alternation in signs. The nature of this alternation is dependent on whether the exponent

• \(n(n+1)/2\)

is odd or even:

• If the exponent is odd, the term is negative.
• If the exponent is even, the term is positive.

This pattern repeats for any substitute value of \(n\), creating a series where signs "alternate" as \(n\) increases: negative, negative, negative, negative, negative. In this specific sequence, due to consistently having an odd exponent, all initial terms evaluated were negative.

Substitution Method

The substitution method is a fundamental technique in algebra and calculus, useful for finding specific terms in sequences or equations by substituting known variables. In sequences, it often involves replacing the sequence variable \(n\) with specific integers to compute the term's value.

In our exercise, substituting values of \(n\) from 1 to 5 into the sequence formula \(a_n = \frac{(-1)^{n(n+1)/2}}{n^2}\) determined each term.

• For \(n=1\), the term calculated is \(-1\).
• For \(n=2\), the term computed is \(-\frac{1}{4}\).
• Continuing with \(n=3\), \(n=4\), and \(n=5\), it results in \(-\frac{1}{9}\), \(-\frac{1}{16}\), and \(-\frac{1}{25}\) respectively.

Each substitution demonstrates the sequence's pattern and the effective use of substitution to determine terms quickly and accurately. This method allows calculation of any term if the formula is known, making it a versatile tool in mathematics.