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Find a formula [tex]\(a_n\)[/tex] for the [tex]\(n\)[/tex]-th term of the following sequence. Assume the series begins at [tex]\(n=1\)[/tex].

[tex]\[
\frac{1}{1}, -\frac{1}{8}, \frac{1}{27}, \ldots
\][/tex]

(Use symbolic notation and fractions where needed.)

[tex]\[
a_n =
\][/tex]

\qquad

Find a formula [tex]\(b_n\)[/tex] for the [tex]\(n\)[/tex]-th term of the following sequence. Assume the series begins at [tex]\(n=1\)[/tex].

[tex]\[
\frac{4}{6}, \frac{5}{7}, \frac{6}{8}, \ldots
\][/tex]

(Use symbolic notation and fractions where needed.)

[tex]\[
b_n =
\][/tex]

[tex]\(\square\)[/tex]


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]