Question

use the explicit formula to find a recursive formula for the sequence $a_n$. write your answer in simplest form.
the recursive formula should depend on $a_{n - 1}$.
$a_n = -4^{n - 1}$
$a_n = \square$

Explanation:

Step1: Recall recursive formula concept

A recursive formula for a sequence relates \( a_n \) to \( a_{n - 1} \). For a geometric sequence, the recursive formula is \( a_n = r \cdot a_{n - 1} \), where \( r \) is the common ratio. First, identify the type of sequence from the explicit formula \( a_n=-4^{n - 1} \).

Step2: Find the common ratio

For the explicit formula of a geometric sequence \( a_n = a_1 \cdot r^{n - 1} \), here \( a_1=-4^{0}=-1 \) (when \( n = 1 \)), and the general term is \( a_n=-4^{n - 1} \). To find the ratio between consecutive terms, compute \( \frac{a_n}{a_{n - 1}} \).

Substitute \( a_n=-4^{n - 1} \) and \( a_{n - 1}=-4^{(n - 1)-1}=-4^{n - 2} \) into the ratio:

$$ \frac{a_n}{a_{n - 1}}=\frac{-4^{n - 1}}{-4^{n - 2}} $$

Using the exponent rule \( \frac{x^m}{x^n}=x^{m - n} \), we get:

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

So the common ratio \( r = 4 \).

Step3: Write the recursive formula

Since it's a geometric sequence with common ratio \( r = 4 \), the recursive formula is \( a_n = 4 \cdot a_{n - 1} \). We also need the initial term. When \( n = 1 \), \( a_1=-4^{1 - 1}=-4^{0}=-1 \). But the problem asks for the recursive formula depending on \( a_{n - 1} \), so the formula is \( a_n = 4a_{n - 1} \) (with \( a_1=-1 \), but the question focuses on the relation between \( a_n \) and \( a_{n - 1} \)).

Answer:

\( 4a_{n - 1} \) (with the understanding that the recursive formula is \( a_n = 4a_{n - 1} \) for \( n \geq 2 \) and \( a_1=-1 \))