To solve this, let's break it down:

Problem

We are given:

• $a_1 = 3$ (the first term of the sequence).
• The recurrence relation: $a_n = a_{n-1} \cdot 6$, which means each term is 6 times the previous one.

Step 1: Recognize the Sequence Type

This is a geometric sequence where:

• The first term ($a_1$) is 3.
• The common ratio ($r$) is 6.

Step 2: Explicit Formula for a Geometric Sequence

The explicit formula for a geometric sequence is: $a_n = a_1 \cdot r^{n-1}$ Substituting the given values ($a_1 = 3$ and $r = 6$): $a_n = 3 \cdot 6^{n-1}$

Final Answer

The explicit formula is: $a_n = 3 \cdot 6^{n-1}$

Would you like further clarification or additional examples?

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Relative Questions:

1. What is the 5th term of the sequence using the explicit formula?
2. How does the explicit formula differ from the recursive formula in this case?
3. What are the characteristics of geometric sequences in general?
4. Can you find the sum of the first $n$ terms of this geometric sequence?
5. How would the formula change if the common ratio were negative?

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Tip:

When transitioning from a recursive formula to an explicit one, always identify whether the sequence is arithmetic or geometric. This simplifies the derivation of the formula!