Given the recurrence relation:

\[
a_n = 2a_{n-1} + 2a_{n-2}
\]

with initial conditions \( a_1 = 5 \) and \( a_2 = 3 \), you can find the subsequent terms \( a_n \) by using the given formula.

Let's calculate the next few terms:

### Calculation for \( a_3 \):
\[
a_3 = 2a_2 + 2a_1 = 2(3) + 2(5) = 6 + 10 = 16
\]

### Calculation for \( a_4 \):
\[
a_4 = 2a_3 + 2a_2 = 2(16) + 2(3) = 32 + 6 = 38
\]

### Calculation for \( a_5 \):
\[
a_5 = 2a_4 + 2a_3 = 2(38) + 2(16) = 76 + 32 = 108
\]

Thus, the sequence \( \{a_n\} \) starts as follows:
- \( a_1 = 5 \)
- \( a_2 = 3 \)
- \( a_3 = 16 \)
- \( a_4 = 38 \)
- \( a_5 = 108 \)

Would you like further details on how to derive these results, or any additional questions?

### Relative Questions:
1. How would you find a general formula for the sequence?
2. Can you determine if this sequence is arithmetic, geometric, or neither?
3. How does the growth rate of this sequence compare to simpler sequences like \(a_n = 2a_{n-1}\)?
4. What would be the value of \(a_n\) if the initial conditions were different, say \(a_1 = 1\) and \(a_2 = 2\)?
5. How does changing the coefficients in the recurrence relation affect the sequence?

### Tip:
When dealing with recurrence relations, identifying patterns in the first few terms can often provide insights into the nature of the sequence and help in formulating a general solution.

Explore how to solve the recurrence relation a_n = 2a_(n-1) + 2a_(n-2) with initial conditions a_1 = 5 and a_2 = 3. This mathematical problem involves calculating subsequent terms and understanding sequence growth patterns. Suitable for Grades 10-12 students studying advanced mathematics.

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