The problem asks to find the formula for the $n$th term of the sequence $-2, 4, -8, 16, \dots$. The possible formulas are: A. $T_n = 2^n$ B. $T_n = (-2)^n$ C. $T_n = (-2n)$ D. $T_n = n^2$

1. Problem Description

The problem asks to find the formula for the $n$th term of the sequence $-2, 4, -8, 16, \dots$. The possible formulas are:

A. $T_n = 2^n$

B. $T_n = (-2)^n$

C. $T_n = (-2n)$

D. $T_n = n^2$

2. Solution Steps

Let's examine each option:

A. If $T_n = 2^n$, then $T_1 = 2^1 = 2$, $T_2 = 2^2 = 4$, $T_3 = 2^3 = 8$, $T_4 = 2^4 = 16$. This sequence is $2, 4, 8, 16, \dots$, which is not the same as the given sequence.

B. If $T_n = (-2)^n$, then $T_1 = (-2)^1 = -2$, $T_2 = (-2)^2 = 4$, $T_3 = (-2)^3 = -8$, $T_4 = (-2)^4 = 16$. This sequence is $-2, 4, -8, 16, \dots$, which is the same as the given sequence.

C. If $T_n = (-2n)$, then $T_1 = (-2)(1) = -2$, $T_2 = (-2)(2) = -4$, $T_3 = (-2)(3) = -6$, $T_4 = (-2)(4) = -8$. This sequence is $-2, -4, -6, -8, \dots$, which is not the same as the given sequence.

D. If $T_n = n^2$, then $T_1 = 1^2 = 1$, $T_2 = 2^2 = 4$, $T_3 = 3^2 = 9$, $T_4 = 4^2 = 16$. This sequence is $1, 4, 9, 16, \dots$, which is not the same as the given sequence.

Therefore, the correct formula is $T_n = (-2)^n$.

3. Final Answer

B. $T_n = (-2)^n$