Answer :
To find the formulas for the sequences given, we can look at the patterns in the sequences and form expressions based on those patterns.
### Sequence [tex]\( a_n \)[/tex]
The sequence is given as:
[tex]\[ \frac{1}{1}, -\frac{1}{8}, \frac{1}{27}, \ldots \][/tex]
To derive the general form [tex]\( a_n \)[/tex] for this sequence, observe the following pattern:
- The denominators are 1, 8, 27, ... which are [tex]\( 1^3, 2^3, 3^3, \ldots \)[/tex].
- The numerators alternate signs and are always 1.
From this, we can see:
- The general form of the denominator is [tex]\( n^3 \)[/tex].
- The sign alternation can be captured by [tex]\((-1)^{n+1}\)[/tex].
Combining both observations, the [tex]\( n \)[/tex]-th term [tex]\( a_n \)[/tex] of the sequence is:
[tex]\[ a_n = \frac{(-1)^{n+1}}{n^3} \][/tex]
### Sequence [tex]\( b_n \)[/tex]
The sequence is given as:
[tex]\[ \frac{4}{6}, \frac{5}{7}, \frac{6}{8}, \ldots \][/tex]
To derive the general form [tex]\( b_n \)[/tex] for this sequence, observe the following pattern:
- The numerators are 4, 5, 6, ... which can be expressed as [tex]\( n + 3 \)[/tex] because each numerator is 3 more than its index.
- The denominators are 6, 7, 8, ... which can be expressed as [tex]\( n + 5 \)[/tex] because each denominator is 5 more than its index.
Combining these observations, the [tex]\( n \)[/tex]-th term [tex]\( b_n \)[/tex] of the sequence is:
[tex]\[ b_n = \frac{n + 3}{n + 5} \][/tex]
### Final Formulas
Thus, the formulas for the given sequences are:
[tex]\[ a_n = \frac{(-1)^{n+1}}{n^3} \][/tex]
[tex]\[ b_n = \frac{n + 3}{n + 5} \][/tex]
### Sequence [tex]\( a_n \)[/tex]
The sequence is given as:
[tex]\[ \frac{1}{1}, -\frac{1}{8}, \frac{1}{27}, \ldots \][/tex]
To derive the general form [tex]\( a_n \)[/tex] for this sequence, observe the following pattern:
- The denominators are 1, 8, 27, ... which are [tex]\( 1^3, 2^3, 3^3, \ldots \)[/tex].
- The numerators alternate signs and are always 1.
From this, we can see:
- The general form of the denominator is [tex]\( n^3 \)[/tex].
- The sign alternation can be captured by [tex]\((-1)^{n+1}\)[/tex].
Combining both observations, the [tex]\( n \)[/tex]-th term [tex]\( a_n \)[/tex] of the sequence is:
[tex]\[ a_n = \frac{(-1)^{n+1}}{n^3} \][/tex]
### Sequence [tex]\( b_n \)[/tex]
The sequence is given as:
[tex]\[ \frac{4}{6}, \frac{5}{7}, \frac{6}{8}, \ldots \][/tex]
To derive the general form [tex]\( b_n \)[/tex] for this sequence, observe the following pattern:
- The numerators are 4, 5, 6, ... which can be expressed as [tex]\( n + 3 \)[/tex] because each numerator is 3 more than its index.
- The denominators are 6, 7, 8, ... which can be expressed as [tex]\( n + 5 \)[/tex] because each denominator is 5 more than its index.
Combining these observations, the [tex]\( n \)[/tex]-th term [tex]\( b_n \)[/tex] of the sequence is:
[tex]\[ b_n = \frac{n + 3}{n + 5} \][/tex]
### Final Formulas
Thus, the formulas for the given sequences are:
[tex]\[ a_n = \frac{(-1)^{n+1}}{n^3} \][/tex]
[tex]\[ b_n = \frac{n + 3}{n + 5} \][/tex]
