Problem 20 - Algebra and Trigonometry

Question

List the first five terms of each sequence. \(\left\\{d_{n}\right\\}=\left\\{(-1)^{n-1}\left(\frac{n}{2 n-1}\right)\right\\}\)

Step-by-Step Solution

Answer

The first five terms of the sequence are: 1, -\( \frac{2}{3} \), \( \frac{3}{5} \), -\( \frac{4}{7} \), \( \frac{5}{9} \).

Step 1 - Set up the formula

The given sequence is defined by the formula: \(\begin\{d_n\} = \left\{ (-1)^{n-1} \left( \frac{n}{2n-1} \right) \right\}\).

Step 2 - Find the first term (\(d_1\))

Substitute \(n = 1\) into the formula: \((-1)^{1-1} \left( \frac{1}{2\cdot1-1} \right)\) which simplifies to \( (-1)^{0} \left( \frac{1}{1} \right) = 1\). So, \(d_1 = 1\).

Step 3 - Find the second term (\(d_2\))

Substitute \(n = 2\) into the formula: \((-1)^{2-1} \left( \frac{2}{2\cdot2-1} \right)\) which simplifies to \( (-1)^{1} \left( \frac{2}{3} \right) = - \frac{2}{3} \). So, \(d_2 = - \frac{2}{3} \).

Step 4 - Find the third term (\(d_3\))

Substitute \(n = 3\) into the formula: \((-1)^{3-1} \left( \frac{3}{2\cdot3-1} \right)\) which simplifies to \( (-1)^{2} \left( \frac{3}{5} \right) = \frac{3}{5} \). So, \(d_3 = \frac{3}{5} \).

Step 5 - Find the fourth term (\(d_4\))

Substitute \(n = 4\) into the formula: \((-1)^{4-1} \left( \frac{4}{2\cdot4-1} \right)\) which simplifies to \( (-1)^{3} \left( \frac{4}{7} \right) = - \frac{4}{7} \). So, \(d_4 = - \frac{4}{7} \).

Step 6 - Find the fifth term (\(d_5\))

Substitute \(n = 5\) into the formula: \((-1)^{5-1} \left( \frac{5}{2\cdot5-1} \right)\) which simplifies to \( (-1)^{4} \left( \frac{5}{9} \right) = \frac{5}{9} \). So, \(d_5 = \frac{5}{9} \).

Key Concepts

Alternating Sequences

An alternating sequence switches between positive and negative values; this characteristic helps identify the nature of the sequence quickly. In our given formula, the term (-1)^{n-1} generates alternating positive and negative signs:
* When n is odd, (-1)^{n-1} = 1 (positive).
* When n is even, (-1)^{n-1} = -1 (negative).
This structure indicates that terms will alternate in sign as n increases. Understanding this alternating nature helps verify if your calculations are correct, and it plays a significant role in the behavior of the sequence.