What is a formula for the nth term of the given sequence? $$ 4,-8,16 \ldots $$ Answer $$ \begin{array}{ll}a_{n}=4\left(-\frac{1}{2} ight)^{n} & a_{n}=4\left(-\frac{1}{2} ight)^{1-n} \ a_{n}=4(-2)^{1-n} & a_{n}=-2(-2)^{n-1}\end{array} $$

Question

What is a formula for the nth term of the given sequence? $$ 4,-8,16 \ldots $$ Answer $$ \begin{array}{ll}a_{n}=4\left(-\frac{1}{2}ight)^{n} & a_{n}=4\left(-\frac{1}{2}ight)^{1-n} \ a_{n}=4(-2)^{1-n} & a_{n}=-2(-2)^{n-1}\end{array} $$

UpStudy AI · Accepted Answer

To find a formula for the $$ n $$th term of the given sequence $$ 4, -8, 16, \ldots $$, we first need to analyze the pattern in the sequence.

1. **Identify the terms:**
   - The first term $$ a_1 = 4 $$
   - The second term $$ a_2 = -8 $$
   - The third term $$ a_3 = 16 $$

2. **Determine the pattern:**
   - The sequence alternates in sign and appears to be multiplied by a factor of $$-2$$ for each subsequent term:
     - $$ a_2 = 4 \times (-2) = -8 $$
     - $$ a_3 = -8 \times (-2) = 16 $$

3. **Generalize the pattern:**
   - The first term is $$ 4 $$.
   - Each term can be expressed as $$ 4 \times (-2)^{n-1} $$ where $$ n $$ is the term number:
     - For $$ n = 1 $$: $$ a_1 = 4 \times (-2)^{1-1} = 4 $$
     - For $$ n = 2 $$: $$ a_2 = 4 \times (-2)^{2-1} = 4 \times (-2) = -8 $$
     - For $$ n = 3 $$: $$ a_3 = 4 \times (-2)^{3-1} = 4 \times 4 = 16 $$

Thus, the formula for the $$ n $$th term of the sequence is:

$$
a_n = 4 \times (-2)^{n-1}
$$

Now, let's check the provided options:

- $$ a_n = 4\left(-\frac{1}{2}\right)^{n} $$ (not correct)
- $$ a_n = 4\left(-\frac{1}{2}\right)^{1-n} $$ (not correct)
- $$ a_n = 4(-2)^{1-n} $$ (not correct)
- $$ a_n = -2(-2)^{n-1} $$ (this is equivalent to $$ 4(-2)^{n-1} $$)

The correct formula for the $$ n $$th term of the sequence is:

$$
a_n = 4(-2)^{n-1}
$$

This matches the last option when simplified.
The formula for the $$ n $$th term of the sequence is $$ a_n = 4(-2)^{n-1} $$.

What is a formula for the nth term of the given sequence? \( 4,-8,16 \ldots \) Answer \( \begin{array}{ll}a_{n}=4\left(-\frac{1}{2}\right)^{n} & a_{n}=4\left(-\frac{1}{2}\right)^{1-n} \\ a_{n}=4(-2)^{1-n} & a_{n}=-2(-2)^{n-1}\end{array} \)

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