What is a formula for the nth term of the given sequence? $$ \begin{array}{ll}4,2,1 \ldots\end{array} $$ Answer $$ \begin{array}{ll}a_{n}=8\left(\frac{1}{2}\right)^{-n} & a_{n}=4\left(\frac{1}{2}\right)^{n-1} \ a_{n}=4(2)^{n} & a_{n}=8(2)^{n-1}\end{array} $$

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To find a formula for the $$ n $$th term of the given sequence $$ 4, 2, 1, \ldots $$, we first need to identify the pattern in the sequence.

1. **Identify the terms:**
   - The first term $$ a_1 = 4 $$
   - The second term $$ a_2 = 2 $$
   - The third term $$ a_3 = 1 $$

2. **Determine the relationship between the terms:**
   - The sequence appears to be decreasing. 
   - From $$ 4 $$ to $$ 2 $$, the term is multiplied by $$ \frac{1}{2} $$.
   - From $$ 2 $$ to $$ 1 $$, the term is again multiplied by $$ \frac{1}{2} $$.

3. **Generalize the pattern:**
   - The sequence can be expressed as:
     $$
     a_n = 4 \left( \frac{1}{2} \right)^{n-1}
     $$
   - This formula indicates that each term is $$ 4 $$ multiplied by $$ \frac{1}{2} $$ raised to the power of $$ n-1 $$.

4. **Verify the formula:**
   - For $$ n = 1 $$:
     $$
     a_1 = 4 \left( \frac{1}{2} \right)^{1-1} = 4 \cdot 1 = 4
     $$
   - For $$ n = 2 $$:
     $$
     a_2 = 4 \left( \frac{1}{2} \right)^{2-1} = 4 \cdot \frac{1}{2} = 2
     $$
   - For $$ n = 3 $$:
     $$
     a_3 = 4 \left( \frac{1}{2} \right)^{3-1} = 4 \cdot \frac{1}{4} = 1
     $$

Thus, the formula for the $$ n $$th term of the sequence is:
$$
a_n = 4 \left( \frac{1}{2} \right)^{n-1}
$$

This matches one of the options provided in the answer choices.
The formula for the $$ n $$th term of the sequence is $$ a_n = 4 \left( \frac{1}{2} \right)^{n-1} $$.

What is a formula for the nth term of the given sequence? \[ \begin{array}{ll}4,2,1 \ldots\end{array} \] Answer \( \begin{array}{ll}a_{n}=8\left(\frac{1}{2}\right)^{-n} & a_{n}=4\left(\frac{1}{2}\right)^{n-1} \\ a_{n}=4(2)^{n} & a_{n}=8(2)^{n-1}\end{array} \)

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