To find a recursive formula for the given sequence, we first observe the relationships between consecutive terms.
Let's denote the terms of the sequence as follows:
- \( a_1 = 55 \)
- \( a_2 = 33 \)
- \( a_3 = 99 \)
- \( a_4 = 5 \)
- \( a_5 = 297 \)
- \( a_6 = 25 \)
Now, we examine the relationship between these terms:
- \( a_1 = 55 \)
- \( a_2 = 33 \) can be computed as \( \frac{a_1}{\frac{55}{33}} = \frac{55}{\frac{5}{3}} = 33 \)
- \( a_3 = 99 \) can be computed as \( a_2 \times \frac{99}{33} = 33 \times 3 = 99 \)
- \( a_4 = 5 \) can be computed as \( a_3 \times \frac{5}{99} = 99 \times \frac{5}{99} = 5 \)
- \( a_5 = 297 \) can be computed as \( a_4 \times \frac{297}{5} = 5 \times 59.4 = 297 \)
- \( a_6 = 25 \) can be computed as \( a_5 \times \frac{25}{297} = 297 \times \frac{25}{297} = 25 \)
From this exploration, we can attempt to identify a pattern:
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For even \( n \), the terms are quite straightforward since if we manually calculate:
- \( a_2 = 55 \times \frac{3}{5} \)
- \( a_4 = 55 \times \frac{3^2}{5^2} \)
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For odd \( n \), it appears that:
- \( a_3 = 33 \times 3 \)
- \( a_5 = 5 \times 59.4 \)
This leads us to hypothesize that terms with alternating oscillatory behavior are manipulated by consecutive multiplicative factors.
Based on recalibrating forms, we notice:
- The first operation seems to be linked to some \( \frac{k}{d} = \frac{55}{5}, \frac{33}{3}, \frac{99}{99} \).
We can condense the recursive key idea:
a_n = a_{n-1} \cdot k(n)
with \( k(n) = \text{some ratio indicative} \) based on whether \( n \) is even or odd.
Therefore finally, we quote down:
Recursive Formula:
\[ a_n = \begin{cases} \frac{3}{5} a_{n-1} & \text{if } n \text{ is even} \ 3a_{n-2} & \text{if } n \text{ is odd} \end{cases} \]
After cleanly writing it into concise format, it creates a holistic good pattern to iterate through, linking even-odd terms in recursive sequence.
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