Problem 23 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

[\frac{1}{2}, \frac{-1}{6}, \frac{1}{12}, \frac{-1}{20}, \frac{1}{30}]

Step 1: Identify the sequence formula

The given sequence formula is \( t_{n} = \frac{(-1)^{n}}{(n+1)(n+2)} \).

Step 2: Calculate the first term \( t_{0} \)

Substitute \( n = 0 \) into the sequence formula: \( t_{0} = \frac{(-1)^{0}}{(0+1)(0+2)} = \frac{1}{1 \times 2} = \frac{1}{2} \).

Step 3: Calculate the second term \( t_{1} \)

Substitute \( n = 1 \) into the sequence formula: \( t_{1} = \frac{(-1)^{1}}{(1+1)(1+2)} = \frac{-1}{2 \times 3} = \frac{-1}{6} \).

Step 4: Calculate the third term \( t_{2} \)

Substitute \( n = 2 \) into the sequence formula: \( t_{2} = \frac{(-1)^{2}}{(2+1)(2+2)} = \frac{1}{3 \times 4} = \frac{1}{12} \).

Step 5: Calculate the fourth term \( t_{3} \)

Substitute \( n = 3 \) into the sequence formula: \( t_{3} = \frac{(-1)^{3}}{(3+1)(3+2)} = \frac{-1}{4 \times 5} = \frac{-1}{20} \).

Step 6: Calculate the fifth term \( t_{4} \)

Substitute \( n = 4 \) into the sequence formula: \( t_{4} = \frac{(-1)^{4}}{(4+1)(4+2)} = \frac{1}{5 \times 6} = \frac{1}{30} \).

Key Concepts

Sequence Formula

A sequence is a list of numbers in a specific order, defined by a particular formula. In this case, the sequence formula given is:
\( t_{n} = \frac{(-1)^{n}}{(n+1)(n+2)} \).
This formula defines how each term in the sequence is calculated based on its position.
You can see that the formula includes an alternating sign, produced by \( (-1)^{n} \), and a fraction term that depends on n.
This formula allows us to find each term by substituting different values for 'n'.

Term Calculation

To calculate a term in a sequence, you need to replace 'n' in the formula with the specific term number starting from 0.
Below are the steps and calculations for the first five terms:

• First Term (\( t_{0} \)):
  
  Substitute n = 0 into the formula: \( t_{0} = \frac{(-1)^{0}}{(0+1)(0+2)} = \frac{1}{1 \times 2} = \frac{1}{2} \).


• Second Term (\( t_{1} \)):
  
  Substitute n = 1 into the formula: \( t_{1} = \frac{(-1)^{1}}{(1+1)(1+2)} = \frac{-1}{2 \times 3} = \frac{-1}{6} \).


• Third Term (\( t_{2} \)):
  
  Substitute n = 2 into the formula: \( t_{2} = \frac{(-1)^{2}}{(2+1)(2+2)} = \frac{1}{3 \times 4} = \frac{1}{12} \).


• Fourth Term (\( t_{3} \)):
  
  Substitute n = 3 into the formula: \( t_{3} = \frac{(-1)^{3}}{(3+1)(3+2)} = \frac{-1}{4 \times 5} = \frac{-1}{20} \).


• Fifth Term (\( t_{4} \)):
  
  Substitute n = 4 into the formula: \( t_{4} = \frac{(-1)^{4}}{(4+1)(4+2)} = \frac{1}{5 \times 6} = \frac{1}{30} \).

Each term involves simple substitution and arithmetic operations.

Alternating Series

An alternating series is a sequence where the terms change sign in a regular pattern.
This is achieved using \( (-1)^{n} \) in the sequence formula.
When 'n' is even (e.g., 0, 2, 4), \( (-1)^{n} \) becomes 1, making the term positive.
When 'n' is odd (e.g., 1, 3), \( (-1)^{n} \) becomes -1, making the term negative.
The given sequence: \( t_{n} = \frac{(-1)^{n}}{(n+1)(n+2)} \), is an example of an alternating series.
This results in a pattern of signs like positive, negative, positive, and so forth.
Such series are useful in various mathematical contexts, including solving problems in calculus and understanding behavior in certain functions.