Question

question what is a formula for the nth term of the given sequence? 4, 16, 64... answer $a_n = 4(4)^{1 - n}$ $a_n = 1(\frac{1}{4})^{-n}$ $a_n = 4(4)^{n}$ $a_n = 1(4)^{-n}$

Explanation:

Step1: Identify the type of sequence

This is a geometric sequence since $\frac{16}{4}=4$ and $\frac{64}{16} = 4$, with a common - ratio $r = 4$. The first - term $a_1=4$.

Step2: Recall the formula for the nth term of a geometric sequence

The formula for the nth term of a geometric sequence is $a_n=a_1r^{n - 1}$, where $a_1$ is the first term and $r$ is the common ratio.

Step3: Substitute the values of $a_1$ and $r$ into the formula

Here, $a_1 = 4$ and $r = 4$. So $a_n=4\times4^{n - 1}$. Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $a_n=4^{1+(n - 1)}=4^n=4\times4^{n - 1}$.

Answer:

$a_n = 4(4)^{n - 1}$ (equivalent to $a_n = 4^n$), and among the given options, the correct one is $a_n = 4(4)^n$ is incorrect, $a_n = 4(4)^{1 - n}=\frac{4}{4^{n - 1}}$ is incorrect, $a_n = 1(\frac{1}{4})^{-n}=4^n$ is correct, $a_n = 1(4)^{-n}=\frac{1}{4^n}$ is incorrect. So the answer is $a_n = 1(\frac{1}{4})^{-n}$.