Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Let's find formulas for the sequences step-by-step:
### Sequence [tex]\(a_n\)[/tex]
Given sequence: [tex]\(\frac{1}{1}, -\frac{1}{8}, \frac{1}{27}, \ldots\)[/tex]
1. Identify the numerators and denominators separately:
- Numerators: [tex]\(1, -1, 1, \ldots\)[/tex]
- Notice the signs alternate: [tex]\(1, -1, 1, \ldots\)[/tex]
- This can be represented by the term [tex]\((-1)^{n+1}\)[/tex], because for [tex]\(n=1\)[/tex], it gives [tex]\(1\)[/tex]; for [tex]\(n=2\)[/tex], it gives [tex]\(-1\)[/tex]; for [tex]\(n=3\)[/tex], it gives [tex]\(1\)[/tex], and so on.
- Denominators: [tex]\(1, 8, 27, \ldots\)[/tex]
- Notice these are cubes of positive integers: [tex]\(1^3, 2^3, 3^3, \ldots\)[/tex]
2. Combine the patterns:
- The general term for the numerator is [tex]\((-1)^{n+1}\)[/tex].
- The general term for the denominator is [tex]\(n^3\)[/tex].
Therefore, the formula for the [tex]\(n\)[/tex]-th term [tex]\(a_n\)[/tex] is:
[tex]\[ a_n = \frac{(-1)^{n+1}}{n^3} \][/tex]
### Sequence [tex]\(b_n\)[/tex]
Given sequence: [tex]\(\frac{4}{6}, \frac{5}{7}, \frac{6}{8}, \ldots\)[/tex]
1. Identify the numerators and denominators separately:
- Numerators: [tex]\(4, 5, 6, \ldots\)[/tex]
- This is an arithmetic sequence where each term increases by 1 starting from 4.
- The general term for the numerator can be represented as [tex]\(n + 3\)[/tex].
- Denominators: [tex]\(6, 7, 8, \ldots\)[/tex]
- This is another arithmetic sequence where each term increases by 1 starting from 6.
- The general term for the denominator can be represented as [tex]\(n + 5\)[/tex].
2. Combine the patterns:
- The general term for the numerator is [tex]\(n + 3\)[/tex].
- The general term for the denominator is [tex]\(n + 5\)[/tex].
Therefore, the formula for the [tex]\(n\)[/tex]-th term [tex]\(b_n\)[/tex] is:
[tex]\[ b_n = \frac{n + 3}{n + 5} \][/tex]
### Final Formulas
- For the sequence [tex]\(a_n\)[/tex]:
[tex]\[ a_n = \frac{(-1)^{n+1}}{n^3} \][/tex]
- For the sequence [tex]\(b_n\)[/tex]:
[tex]\[ b_n = \frac{n + 3}{n + 5} \][/tex]
Thus, the formulas are:
[tex]\[ \boxed{a_n = \frac{(-1)^{n+1}}{n^3}} \][/tex]
[tex]\[ \boxed{b_n = \frac{n + 3}{n + 5}} \][/tex]
### Sequence [tex]\(a_n\)[/tex]
Given sequence: [tex]\(\frac{1}{1}, -\frac{1}{8}, \frac{1}{27}, \ldots\)[/tex]
1. Identify the numerators and denominators separately:
- Numerators: [tex]\(1, -1, 1, \ldots\)[/tex]
- Notice the signs alternate: [tex]\(1, -1, 1, \ldots\)[/tex]
- This can be represented by the term [tex]\((-1)^{n+1}\)[/tex], because for [tex]\(n=1\)[/tex], it gives [tex]\(1\)[/tex]; for [tex]\(n=2\)[/tex], it gives [tex]\(-1\)[/tex]; for [tex]\(n=3\)[/tex], it gives [tex]\(1\)[/tex], and so on.
- Denominators: [tex]\(1, 8, 27, \ldots\)[/tex]
- Notice these are cubes of positive integers: [tex]\(1^3, 2^3, 3^3, \ldots\)[/tex]
2. Combine the patterns:
- The general term for the numerator is [tex]\((-1)^{n+1}\)[/tex].
- The general term for the denominator is [tex]\(n^3\)[/tex].
Therefore, the formula for the [tex]\(n\)[/tex]-th term [tex]\(a_n\)[/tex] is:
[tex]\[ a_n = \frac{(-1)^{n+1}}{n^3} \][/tex]
### Sequence [tex]\(b_n\)[/tex]
Given sequence: [tex]\(\frac{4}{6}, \frac{5}{7}, \frac{6}{8}, \ldots\)[/tex]
1. Identify the numerators and denominators separately:
- Numerators: [tex]\(4, 5, 6, \ldots\)[/tex]
- This is an arithmetic sequence where each term increases by 1 starting from 4.
- The general term for the numerator can be represented as [tex]\(n + 3\)[/tex].
- Denominators: [tex]\(6, 7, 8, \ldots\)[/tex]
- This is another arithmetic sequence where each term increases by 1 starting from 6.
- The general term for the denominator can be represented as [tex]\(n + 5\)[/tex].
2. Combine the patterns:
- The general term for the numerator is [tex]\(n + 3\)[/tex].
- The general term for the denominator is [tex]\(n + 5\)[/tex].
Therefore, the formula for the [tex]\(n\)[/tex]-th term [tex]\(b_n\)[/tex] is:
[tex]\[ b_n = \frac{n + 3}{n + 5} \][/tex]
### Final Formulas
- For the sequence [tex]\(a_n\)[/tex]:
[tex]\[ a_n = \frac{(-1)^{n+1}}{n^3} \][/tex]
- For the sequence [tex]\(b_n\)[/tex]:
[tex]\[ b_n = \frac{n + 3}{n + 5} \][/tex]
Thus, the formulas are:
[tex]\[ \boxed{a_n = \frac{(-1)^{n+1}}{n^3}} \][/tex]
[tex]\[ \boxed{b_n = \frac{n + 3}{n + 5}} \][/tex]
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
