Answer :
To determine the recursive formula for the given sequence [tex]\(5, 35, 245, 1,715, 12,005, 84,035, \ldots\)[/tex], let's analyze how the sequence progresses.
Step-by-step solution:
1. Identify consecutive terms:
- The sequence starts with: 5, 35, 245, 1715, 12005, 84035.
2. Calculate the ratios between consecutive terms:
- To see if there's a pattern, we calculate the ratio of each term to the previous one:
- [tex]\( \frac{35}{5} = 7 \)[/tex]
- [tex]\( \frac{245}{35} = 7 \)[/tex]
- [tex]\( \frac{1715}{245} = 7 \)[/tex]
- [tex]\( \frac{12005}{1715} = 7 \)[/tex]
- [tex]\( \frac{84035}{12005} = 7 \)[/tex]
3. Check consistency:
- All the ratios are equal to 7. This indicates that the sequence is a geometric progression where each term is 7 times the previous term.
4. Recursive formula:
- For a geometric sequence where the ratio between consecutive terms is constant, the recursive formula can be written as:
[tex]\[
f(n) = 7 \cdot f(n-1) \quad \text{for} \ n > 1
\][/tex]
- Our starting term [tex]\( f(1) = 5 \)[/tex], and each subsequent term is 7 times the previous one.
Thus, the recursive formula that defines this sequence for [tex]\( n > 1 \)[/tex] is:
[tex]\[ f(n) = 7 \cdot f(n-1) \][/tex]
This formula matches with the progression observed in the sequence.
Step-by-step solution:
1. Identify consecutive terms:
- The sequence starts with: 5, 35, 245, 1715, 12005, 84035.
2. Calculate the ratios between consecutive terms:
- To see if there's a pattern, we calculate the ratio of each term to the previous one:
- [tex]\( \frac{35}{5} = 7 \)[/tex]
- [tex]\( \frac{245}{35} = 7 \)[/tex]
- [tex]\( \frac{1715}{245} = 7 \)[/tex]
- [tex]\( \frac{12005}{1715} = 7 \)[/tex]
- [tex]\( \frac{84035}{12005} = 7 \)[/tex]
3. Check consistency:
- All the ratios are equal to 7. This indicates that the sequence is a geometric progression where each term is 7 times the previous term.
4. Recursive formula:
- For a geometric sequence where the ratio between consecutive terms is constant, the recursive formula can be written as:
[tex]\[
f(n) = 7 \cdot f(n-1) \quad \text{for} \ n > 1
\][/tex]
- Our starting term [tex]\( f(1) = 5 \)[/tex], and each subsequent term is 7 times the previous one.
Thus, the recursive formula that defines this sequence for [tex]\( n > 1 \)[/tex] is:
[tex]\[ f(n) = 7 \cdot f(n-1) \][/tex]
This formula matches with the progression observed in the sequence.
