Given the first term $a_1 = 5$ and the recursion formula:

$a_{n+1} = \frac{n \cdot a_n}{n + 3}$

we can find the first ten terms of the sequence by calculating each term based on the previous one.

Step-by-Step Calculation of Each Term

1. First term: $a_1 = 5$
2. Second term: $a_2 = \frac{1 \cdot a_1}{1 + 3} = \frac{1 \cdot 5}{4} = \frac{5}{4} = 1.25$
3. Third term: $a_3 = \frac{2 \cdot a_2}{2 + 3} = \frac{2 \cdot 1.25}{5} = \frac{2.5}{5} = 0.5$
4. Fourth term: $a_4 = \frac{3 \cdot a_3}{3 + 3} = \frac{3 \cdot 0.5}{6} = \frac{1.5}{6} = 0.25$
5. Fifth term: $a_5 = \frac{4 \cdot a_4}{4 + 3} = \frac{4 \cdot 0.25}{7} = \frac{1}{7} \approx 0.1429$
6. Sixth term: $a_6 = \frac{5 \cdot a_5}{5 + 3} = \frac{5 \cdot 0.1429}{8} = \frac{0.7145}{8} \approx 0.0893$
7. Seventh term: $a_7 = \frac{6 \cdot a_6}{6 + 3} = \frac{6 \cdot 0.0893}{9} = \frac{0.5358}{9} \approx 0.0595$
8. Eighth term: $a_8 = \frac{7 \cdot a_7}{7 + 3} = \frac{7 \cdot 0.0595}{10} = \frac{0.4165}{10} \approx 0.0417$
9. Ninth term: $a_9 = \frac{8 \cdot a_8}{8 + 3} = \frac{8 \cdot 0.0417}{11} = \frac{0.3336}{11} \approx 0.0303$
10. Tenth term: $a_{10} = \frac{9 \cdot a_9}{9 + 3} = \frac{9 \cdot 0.0303}{12} = \frac{0.2727}{12} \approx 0.0227$

First Ten Terms of the Sequence

Thus, the first ten terms are approximately: $5, 1.25, 0.5, 0.25, 0.1429, 0.0893, 0.0595, 0.0417, 0.0303, 0.0227$

Would you like more details or have any questions?

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Related Questions:

1. How does the recursion formula affect the behavior of this sequence as $n$ increases?
2. What is the general behavior of the sequence in terms of convergence?
3. How would the sequence change if the initial term $a_1$ were different?
4. Can the recursive formula be expressed as a closed-form expression?
5. How can this type of sequence be applied in real-world contexts?

Tip: Recurrence relations often lead to sequences that either converge, oscillate, or grow. Analyzing the recursion formula can give insights into the behavior of the sequence.