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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]
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