Problem 137
Question
Write the first five terms of the sequence. $$a_{n}=\frac{(-1)^{n} x^{2 n}}{(2 n) !}$$
Step-by-Step Solution
Verified Answer
The first five terms of the sequence are: \(a_{1}=-\frac{x^{2}}{2}\), \(a_{2}=\frac{x^{4}}{24}\), \(a_{3}= -\frac{x^{6}}{720}\), \(a_{4}=\frac{x^{8}}{40320}\), \(a_{5}= -\frac{x^{10}}{3628800}\)
1Step 1: Substitute n=1 into the sequence
When you substitute n=1 into the sequence formula, the equation becomes \(a_{1} = \frac{(-1)^{1} x^{2\cdot1}}{(2\cdot1)!} \) which simplifies to \(a_{1}=-\frac{x^{2}}{2}\)
2Step 2: Substitute n=2 into the sequence
Substitute n=2 into the sequence formula. The equation becomes \(a_{2} = \frac{(-1)^{2} x^{2\cdot2}}{(2\cdot2)!}\) which simplifies to \(a_{2}=\frac{x^{4}}{24}\)
3Step 3: Substitute n=3 into the sequence
When you substitute n=3 into the sequence formula, the equation becomes \(a_{3} = \frac{(-1)^{3} x^{2\cdot3}}{(2\cdot3)!}\) which simplifies to \(a_{3}= -\frac{x^{6}}{720}\)
4Step 4: Substitute n=4 into the sequence
Substitute n=4 into the sequence formula. The equation becomes \(a_{4} = \frac{(-1)^{4} x^{2\cdot4}}{(2\cdot4)!}\) which simplifies to \(a_{4}=\frac{x^{8}}{40320}\)
5Step 5: Substitute n=5 into the sequence
When you substitute n=5 into the sequence formula, the equation becomes \(a_{5} = \frac{(-1)^{5} x^{2\cdot5}}{(2\cdot5)!}\) which simplifies to \(a_{5}= -\frac{x^{10}}{3628800}\)
Key Concepts
FactorialAlternating SeriesExponentsTerm Simplification
Factorial
A factorial, denoted by an exclamation mark (!) following a number, is the product of all positive integers up to that number. It's a way of multiplying a sequence of descending natural numbers. Factorials are particularly important in permutations, combinations, and various mathematical sequences. For example:
- 4! = 4 × 3 × 2 × 1 = 24
- 5! = 5 × 4 × 3 × 2 × 1 = 120
Alternating Series
An alternating series is a sequence where the signs alternate between positive and negative. In our sequence, we see this structure from the \((-1)^n\) term. Here's how it affects the sequence:
- When \( n \) is odd, \((-1)^n\) results in a negative sign.
- When \( n \) is even, \((-1)^n\) results in a positive sign.
Exponents
Exponents provide a way to express repeated multiplication of a number by itself. In this sequence, exponents appear in the numerator of the sequence term as \( x^{2n} \). More specifically:
- The base \( x \) is repeatedly multiplied by itself \( 2n \) times.
- For the first term, \( x^2\) represents \( x \) times \( x \).
- In the second term, \( x^4 \) means \( x \times x \times x \times x \).
Term Simplification
Term simplification is the process of making an expression or equation simpler and more readable. We use simplification throughout mathematics to make calculations easier and more intuitive. The given sequence uses simplification in each of its terms to reveal its core structure:
- Initially, replace \( n \) with specific values to calculate each term, such as \( n = 1, 2, 3, \) etc.
- Calculate each component separately: \((-1)^n\), \(x^{2n}\), and \((2n)!\).
- Combine and simplify the calculation step by step to reduce to the simplest fractional form.
Other exercises in this chapter
Problem 135
Write the first five terms of the sequence. $$a_{n}=\frac{(-1)^{n} x^{2 n+1}}{2 n+1}$$
View solution Problem 136
Write the first five terms of the sequence. $$a_{n}=\frac{(-1)^{n} x^{n+1}}{n+1}$$
View solution Problem 138
Write the first five terms of the sequence. $$a_{n}=\frac{(-1)^{n} x^{2 n+1}}{(2 n+1) !}$$
View solution Problem 139
Write the first five terms of the sequence. $$a_{n}=\frac{(-1)^{n} x^{n}}{n !}$$
View solution