Chapter 11: Problem 33
Find the first five terms of the infinite sequence whose nth term is given. $$c_{n}=\frac{(-1)^{n}}{(n+1) !}$$
Short Answer
Expert verified
The first five terms are -\frac{1}{2}, \frac{1}{6}, -\frac{1}{24}, \frac{1}{120}, and -\frac{1}{720}.
Step by step solution
01
Understand the Given Formula
The exercise gives the formula for the nth term of a sequence: \[ c_n = \frac{(-1)^n}{(n+1)!} \]Here, 'n' represents the term number in the sequence. '!' represents a factorial, which is the product of all positive integers up to that number.
02
Calculate the First Term (c_1)
Substitute n = 1 into the formula: \[ c_1 = \frac{(-1)^1}{(1+1)!} = \frac{-1}{2!} = \frac{-1}{2} Therefore, the first term (c_1) is -\frac{1}{2}. \]
03
Calculate the Second Term (c_2)
Substitute n = 2 into the formula: \[ c_2 = \frac{(-1)^2}{(2+1)!} = \frac{1}{3!} = \frac{1}{6} Therefore, the second term (c_2) is \frac{1}{6}. \]
04
Calculate the Third Term (c_3)
Substitute n = 3 into the formula: \[ c_3 = \frac{(-1)^3}{(3+1)!} = \frac{-1}{4!} = \frac{-1}{24} Therefore, the third term (c_3) is -\frac{1}{24}. \]
05
Calculate the Fourth Term (c_4)
Substitute n = 4 into the formula: \[ c_4 = \frac{(-1)^4}{(4+1)!} = \frac{1}{5!} = \frac{1}{120} Therefore, the fourth term (c_4) is \frac{1}{120}. \]
06
Calculate the Fifth Term (c_5)
Substitute n = 5 into the formula: \[ c_5 = \frac{(-1)^5}{(5+1)!} = \frac{-1}{6!} = \frac{-1}{720} Therefore, the fifth term (c_5) is -\frac{1}{720}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
infinite sequence
An infinite sequence is a set of numbers arranged in a specific order, without any end. Unlike a finite sequence, which has a predefined number of terms, an infinite sequence goes on forever. Each number in the sequence is called a term. For instance, the sequence defined by the formula \( c_n = \frac{(-1)^n}{(n+1)!} \) generates an infinite number of terms as you keep substituting higher and higher values for 'n'.
- The notation c\textsubscript{n} represents the nth term of the sequence.
- The sequence continues indefinitely, meaning there's no last term.
factorial
The factorial of a number, denoted as n!, is the product of all positive integers up to that number. For example, 4! (read as 'four factorial') is calculated as 4 \times 3 \times 2 \times 1 = 24.
- 0! is defined as 1, because the product of an empty set is 1 by convention.
- Factorials grow very quickly; for instance, 5! = 120 and 6! = 720.
alternating series
An alternating series is a sequence where the terms alternate in sign. This means that the signs of the terms switch back and forth between positive and negative. The given sequence \( c_n = \frac{(-1)^n}{(n+1)!} \) is an example.
- When n is odd, \( (-1)^n \) gives a negative term.
- When n is even, \( (-1)^n \) gives a positive term.
nth term
The nth term of a sequence is a formula that allows you to find the value of any term in the sequence based on its position number, n. In our exercise, the nth term is given by \( c_n = \frac{(-1)^n}{(n+1)!} \).
- By substituting different values of n into the formula, you can calculate specific terms.
- For example, when n=1, we have \( c_1 = \frac{(-1)^1}{(1+1)!} = \frac{-1}{2} \).