Question

use the initial term and the recursive formula to find an explicit formula for the sequence $a_n$. write your answer in simplest form. $a_1 = -1$ $a_n = -3a_{n - 1}$ $a_n = \square$

Explanation:

Step1: Identify sequence type

This is a geometric sequence, where each term is a constant multiple of the previous term. The recursive formula $a_n = -3a_{n-1}$ tells us the common ratio $r = -3$, and the initial term $a_1 = -1$.

Step2: Recall geometric sequence formula

The explicit formula for a geometric sequence starting at $n=1$ is $a_n = a_1 \cdot r^{n-1}$.

Step3: Substitute known values

Substitute $a_1 = -1$ and $r = -3$ into the formula:
$a_n = (-1) \cdot (-3)^{n-1}$

Step4: Simplify the expression

We can rewrite $(-1) \cdot (-3)^{n-1}$ as $(-3)^{n-1} \cdot (-1) = (-3)^{n-1} \cdot (-3)^0 \cdot (-1) = (-3)^n \cdot \frac{-1}{-3}$? No, simpler simplification: $(-1) \cdot (-3)^{n-1} = (-1) \cdot \frac{(-3)^n}{-3} = \frac{(-3)^n}{3}$? Wait, better to combine signs:
$(-1) \cdot (-3)^{n-1} = (-1) \cdot (-1)^{n-1} \cdot 3^{n-1} = (-1)^n \cdot 3^{n-1}$
Or alternatively:
$(-1) \cdot (-3)^{n-1} = (-3)^{n-1} \times (-1) = (-3)^{n-1} \times (-3)^0 \times (-1) = (-3)^n \times \frac{-1}{-3}$ no, correct simplification: $(-1) \times (-3)^{n-1} = (-3)^{n-1} \times (-1) = (-1)^{n-1}3^{n-1} \times (-1) = (-1)^n 3^{n-1}$
Or we can write it as $a_n = -(-3)^{n-1}$ which is equivalent, but the simplest form is combining the constants:
$(-1) \times (-3)^{n-1} = (-1) \times \frac{(-3)^n}{-3} = \frac{(-3)^n}{3} = (-3)^n \times 3^{-1} = (-1)^n 3^{n-1}$

Answer:

$a_n = (-1)^n 3^{n-1}$ or equivalently $a_n = -(-3)^{n-1}$ (both are simplified, but $(-1)^n 3^{n-1}$ is a clean explicit form)