2. What are the first five terms of the recursive sequence? $$ a_{n}=3 a_{n-1}+3 $$ where

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We are given the recurrence
$$
a_{n}=3a_{n-1}+3.
$$
Because the initial term is not specified, we will express the first five terms in terms of an initial value, which we denote by $$ a_1 $$.

**Step 1. Write the first term.**

The first term is simply the initial value:
$$
a_1 \quad \text{(given)}.
$$

**Step 2. Compute the second term.**

Using the recurrence for $$ n=2 $$:
$$
a_2 = 3a_{1}+3.
$$

**Step 3. Compute the third term.**

For $$ n=3 $$:
$$
a_3 = 3a_{2} + 3.
$$
Substitute the expression for $$ a_2 $$:
$$
a_3 = 3\bigl(3a_{1}+3\bigr) + 3 = 9a_{1} + 9 + 3 = 9a_{1} + 12.
$$

**Step 4. Compute the fourth term.**

For $$ n=4 $$:
$$
a_4 = 3a_{3} + 3.
$$
Substitute the expression for $$ a_3 $$:
$$
a_4 = 3\bigl(9a_{1}+12\bigr) + 3 = 27a_{1} + 36 + 3 = 27a_{1} + 39.
$$

**Step 5. Compute the fifth term.**

For $$ n=5 $$:
$$
a_5 = 3a_{4} + 3.
$$
Substitute the expression for $$ a_4 $$:
$$
a_5 = 3\bigl(27a_{1}+39\bigr) + 3 = 81a_{1} + 117 + 3 = 81a_{1} + 120.
$$

Thus, the first five terms of the sequence expressed in terms of the initial value $$ a_1 $$ are:
$$
\begin{aligned}
a_1 &= a_1,\[1mm]
a_2 &= 3a_{1}+3,\[1mm]
a_3 &= 9a_{1}+12,\[1mm]
a_4 &= 27a_{1}+39,\[1mm]
a_5 &= 81a_{1}+120.
\end{aligned}
$$

If a specific numerical initial value is provided (for example, if $$ a_1 = 3 $$), you can substitute that value into these expressions to obtain numerical answers.
The first five terms of the sequence are: 1. $$ a_1 = a_1 $$ 2. $$ a_2 = 3a_1 + 3 $$ 3. $$ a_3 = 9a_1 + 12 $$ 4. $$ a_4 = 27a_1 + 39 $$ 5. $$ a_5 = 81a_1 + 120 $$

What are the first five terms of the recursive sequence?

where

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What are the first five terms of the recursive sequence? where