Problem 97
Question
The sequence \(\mid a_{n}\) \\} is recursively defined. Compute \(a_{n}\) for \(n=1,2, \ldots, 5\) $$ a_{n+1}=\frac{a_{n}}{1+a_{n}}, a_{0}=1 $$
Step-by-Step Solution
Verified Answer
The terms are \( a_1 = \frac{1}{2} \), \( a_2 = \frac{1}{3} \), \( a_3 = \frac{1}{4} \), \( a_4 = \frac{1}{5} \), and \( a_5 = \frac{1}{6} \).
1Step 1: Calculate a_1
Start with the initial value, \( a_0 = 1 \). Using the recursive formula, compute \( a_1 \):\[ a_1 = \frac{a_0}{1 + a_0} = \frac{1}{1 + 1} = \frac{1}{2}. \]
2Step 2: Calculate a_2
Using the value of \( a_1 \) from the previous step, compute \( a_2 \):\[ a_2 = \frac{a_1}{1 + a_1} = \frac{\frac{1}{2}}{1 + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3}. \]
3Step 3: Calculate a_3
Using the value of \( a_2 \), compute \( a_3 \):\[ a_3 = \frac{a_2}{1 + a_2} = \frac{\frac{1}{3}}{1 + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{4}{3}} = \frac{1}{4}. \]
4Step 4: Calculate a_4
Using the value of \( a_3 \), compute \( a_4 \):\[ a_4 = \frac{a_3}{1 + a_3} = \frac{\frac{1}{4}}{1 + \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{5}{4}} = \frac{1}{5}. \]
5Step 5: Calculate a_5
Using the value of \( a_4 \), compute \( a_5 \):\[ a_5 = \frac{a_4}{1 + a_4} = \frac{\frac{1}{5}}{1 + \frac{1}{5}} = \frac{\frac{1}{5}}{\frac{6}{5}} = \frac{1}{6}. \]
Key Concepts
Understanding the Recursive FormulaEstablishing the Initial ValueCarrying Out Sequence Computation
Understanding the Recursive Formula
A recursive formula is a way to determine the elements of a sequence using previous elements. It's like following a set of instructions where the result of one step leads to the next. In mathematics, recursive formulas are often used to model situations where each new term depends on one or more preceding terms.
For our given problem, the recursive formula is:
For our given problem, the recursive formula is:
- This formula describes how to find \(a_{n+1}\), the next term in the sequence, by using \(a_n\), the current term.
- The operation \(\frac{a_n}{1 + a_n}\) means we divide the current term \(a_n\) by one more than itself.
Establishing the Initial Value
The initial value of a sequence is the starting point from which all other elements are calculated using the recursive formula. It's essential because without a starting point, the recursive formula cannot generate any subsequent terms. In our exercise, the initial value given is:
- \(a_0 = 1\)
Carrying Out Sequence Computation
Sequence computation is the process of determining all the terms in a sequence by using the recursive formula and initial value. Each term is derived from the previous one following the strict rule established by the recursive formula. Let's see how the sequence is computed from \(n=1\) to \(n=5\) utilizing our recursive formula:
- For \(a_1\): Start with \(a_0 = 1\), and apply the formula: \(a_1 = \frac{1}{1 + 1} = \frac{1}{2}\)
- For \(a_2\): Use \(a_1 = \frac{1}{2}\), so: \(a_2 = \frac{\frac{1}{2}}{1 + \frac{1}{2}} = \frac{1}{3}\)
- For \(a_3\): With \(a_2 = \frac{1}{3}\), compute: \(a_3 = \frac{\frac{1}{3}}{1 + \frac{1}{3}} = \frac{1}{4}\)
- For \(a_4\): Given \(a_3 = \frac{1}{4}\), find: \(a_4 = \frac{\frac{1}{4}}{1 + \frac{1}{4}} = \frac{1}{5}\)
- For \(a_5\): Using \(a_4 = \frac{1}{5}\), calculate: \(a_5 = \frac{\frac{1}{5}}{1 + \frac{1}{5}} = \frac{1}{6}\)
Other exercises in this chapter
Problem 95
The sequence \(\mid a_{n}\) \\} is recursively defined. Compute \(a_{n}\) for \(n=1,2, \ldots, 5\) $$ a_{n+1}=1+2 a_{n}, a_{0}=0 $$
View solution Problem 96
The sequence \(\mid a_{n}\) \\} is recursively defined. Compute \(a_{n}\) for \(n=1,2, \ldots, 5\) $$ a_{n+1}=4-2 a_{n}, a_{0}=\frac{4}{3} $$
View solution Problem 98
The sequence \(\mid a_{n}\) \\} is recursively defined. Compute \(a_{n}\) for \(n=1,2, \ldots, 5\) $$ a_{n+1}=\sqrt{a_{n}}, a_{0}=16 $$
View solution Problem 99
The sequence \(\mid a_{n}\) \\} is recursively defined. Compute \(a_{n}\) for \(n=1,2, \ldots, 5\) $$ a_{n+1}=a_{n}+\frac{1}{a_{n}}, a_{0}=1 $$
View solution