To write an explicit formula for the given sequence, we analyze the pattern observed in the terms.

Sequence Recap:

The sequence is: $\frac{1}{8}, \frac{1}{32}, \frac{1}{128}, \frac{1}{512}, \dots$

Recursive Formula:

The recursive formula is: $A_n = \frac{1}{4} \times A_{n-1}$

Deriving the Explicit Formula:

The pattern shows that each term is multiplied by $\frac{1}{4}$ compared to the previous term. This indicates a geometric sequence.

For a geometric sequence, the general explicit formula is: $A_n = A_1 \times r^{n-1}$

Where:

• $A_1$ is the first term (here $\frac{1}{8}$)
• $r$ is the common ratio (here $\frac{1}{4}$)

So, the explicit formula is: $A_n = \frac{1}{8} \times \left(\frac{1}{4}\right)^{n-1}$

Simplifying the Formula:

$A_n = \frac{1}{8} \times \frac{1}{4^{n-1}}$

$A_n = \frac{1}{8 \times 4^{n-1}}$

Since $4^{n-1} = (2^2)^{n-1} = 2^{2(n-1)}$, the formula can also be expressed as: $A_n = \frac{1}{2^{3 + 2(n-1)}}$

Final Answer:

$A_n = \frac{1}{8 \times 4^{n-1}}$

Would you like more details or have any questions?

Here are 5 related questions:

1. How does this formula change if the first term is different?
2. Can you determine the 10th term using this explicit formula?
3. How does this sequence behave as $n$ approaches infinity?
4. What is the sum of the first $n$ terms of this sequence?
5. How would this formula be affected if the common ratio was $\frac{1}{3}$ instead of $\frac{1}{4}$?

Tip: Converting recursive formulas to explicit formulas helps find any term directly without needing to calculate all previous terms.