To determine the first 8 terms of the sequence defined by the recursive formula:

Formula:

1. $a_1 = 0$
2. $a_2 = 1$
3. $a_3 = 7$
4. $a_n = \frac{a_{n-1}}{a_{n-2}} + a_{n-3}$, for $n \geq 4$.

Step-by-Step Calculation:

Given:

• $a_1 = 0$
• $a_2 = 1$
• $a_3 = 7$

Now, calculate subsequent terms using the recursive formula:

$a_4$:

$a_4 = \frac{a_3}{a_2} + a_1 = \frac{7}{1} + 0 = 7$

$a_5$:

$a_5 = \frac{a_4}{a_3} + a_2 = \frac{7}{7} + 1 = 1 + 1 = 2$

$a_6$:

$a_6 = \frac{a_5}{a_4} + a_3 = \frac{2}{7} + 7 = \frac{2}{7} + \frac{49}{7} = \frac{51}{7}$

$a_7$:

$a_7 = \frac{a_6}{a_5} + a_4 = \frac{\frac{51}{7}}{2} + 7 = \frac{51}{14} + 7 = \frac{51}{14} + \frac{98}{14} = \frac{149}{14}$

$a_8$:

$a_8 = \frac{a_7}{a_6} + a_5 = \frac{\frac{149}{14}}{\frac{51}{7}} + 2 = \frac{149}{14} \cdot \frac{7}{51} + 2 = \frac{149}{102} + 2 = \frac{149}{102} + \frac{204}{102} = \frac{353}{102}$

Final Sequence:

The first 8 terms are: $a_1 = 0, \, a_2 = 1, \, a_3 = 7, \, a_4 = 7, \, a_5 = 2, \, a_6 = \frac{51}{7}, \, a_7 = \frac{149}{14}, \, a_8 = \frac{353}{102}.$

Would you like more explanation on any step?

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Related Questions:

1. Can you find the general behavior or limit of this sequence?
2. What happens if $a_1$, $a_2$, or $a_3$ are changed?
3. How does this sequence relate to convergent/divergent series?
4. Can we analyze $a_n$ as $n \to \infty$?
5. Can this be implemented programmatically for larger values of $n$?

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Tip: For sequences with recursive definitions, substituting terms carefully and simplifying fractions will help avoid calculation errors.