Problem 12 - Precalculus: Mathematics for Calculus

Question

Find the first four terms and the 100 th term of the sequence whose \(n\)th term is given. \(a_{n}=(-1)^{n+1} \frac{n}{n+1}\)

Step-by-Step Solution

Answer

The first four terms are \(\frac{1}{2}, -\frac{2}{3}, \frac{3}{4}, -\frac{4}{5}\), and the 100th term is \(-\frac{100}{101}\).

Step 1: Identify the Formula

The given sequence is defined by the formula \(a_{n}=(-1)^{n+1} \frac{n}{n+1}\). This formula will help us calculate any term of the sequence by substituting the value of \(n\) into it.

Step 2: Calculate the First Term

To find the first term, substitute \(n=1\) into the formula: \[ a_{1} = (-1)^{1+1} \frac{1}{1+1} = (-1)^{2} \frac{1}{2} = 1 \times \frac{1}{2} = \frac{1}{2}. \] So the first term \(a_1\) is \(\frac{1}{2}\).

Step 3: Calculate the Second Term

Substitute \(n=2\) into the formula:\[ a_{2} = (-1)^{2+1} \frac{2}{2+1} = (-1)^{3} \frac{2}{3} = -1 \times \frac{2}{3} = -\frac{2}{3}. \] So the second term \(a_2\) is \(-\frac{2}{3}\).

Step 4: Calculate the Third Term

Substitute \(n=3\) into the formula:\[ a_{3} = (-1)^{3+1} \frac{3}{3+1} = (-1)^{4} \frac{3}{4} = 1 \times \frac{3}{4} = \frac{3}{4}. \] So the third term \(a_3\) is \(\frac{3}{4}\).

Step 5: Calculate the Fourth Term

Substitute \(n=4\) into the formula:\[ a_{4} = (-1)^{4+1} \frac{4}{4+1} = (-1)^{5} \frac{4}{5} = -1 \times \frac{4}{5} = -\frac{4}{5}. \] So the fourth term \(a_4\) is \(-\frac{4}{5}\).

Step 6: Calculate the 100th Term

Substitute \(n=100\) into the formula:\[ a_{100} = (-1)^{100+1} \frac{100}{100+1} = (-1)^{101} \frac{100}{101} = -1 \times \frac{100}{101} = -\frac{100}{101}. \] So the 100th term \(a_{100}\) is \(-\frac{100}{101}\).

Key Concepts

Arithmetic sequences

When exploring sequences, arithmetic sequences are fundamental. They are formed by adding a constant value, called the "common difference," to each term to get the next one. For example, in the sequence 2, 4, 6, 8, each term increases by 2. This makes it easy to predict any term in the sequence just by knowing the first term and the common difference. Since the common difference remains the same, arithmetic sequences follow a linear pattern. You can use the formula \( a_n = a_1 + (n-1)d \) to find any term in an arithmetic sequence. In this formula, \( a_n \) is the \( n \)-th term, \( a_1 \) is the first term, \( d \) is the common difference, and \( n \) is the term's position in the sequence.

Alternating sequence

An alternating sequence alternates between positive and negative values. The exercise you're working on involves an alternating sequence. This is evident from the given formula \( a_n=(-1)^{n+1}\frac{n}{n+1} \).
This alternate sign pattern is due to the \((-1)^{n+1}\) part of the formula. It creates positive terms when \(n\) is odd and negative terms when \(n\) is even. Such sequences do not follow the arithmetic pattern because their sign changes regularly.
Alternating sequences are useful in various mathematical contexts, including series approximation and trigonometric expansions. Understanding their behavior is key to mastering more complex calculus concepts.

Formula for nth term

The formula for the \(n\)-th term of a sequence is a powerful tool in mathematics. It gives a way to calculate any specific term without needing all the prior terms. In your exercise, the formula is \(a_n=(-1)^{n+1}\frac{n}{n+1} \).
This particular formula has two key components:

• \((-1)^{n+1}\): This regulates the sign of the term, making the sequence alternate between positive and negative.
• \(\frac{n}{n+1}\): This controls the actual numerical value of each term, determining how it changes as \(n\) increases.

Using such a formula, you can efficiently find any term in the sequence. For instance, substituting \(n=100\) directly gives the 100th term. Always remember, understanding the structure of the formula helps a lot in predicting the sequence's behavior.