Question

find the first four terms of the sequence given by the following.
$a_n = 6\cdot (-2)^{n - 1}$, $n = 1, 2, 3...$

Explanation:

Step1: Calculate n=1 term

$a_1 = 6\cdot(-2)^{1-1} = 6\cdot(-2)^0 = 6\cdot1 = 6$

Step2: Calculate n=2 term

$a_2 = 6\cdot(-2)^{2-1} = 6\cdot(-2)^1 = 6\cdot(-2) = -12$

Step3: Calculate n=3 term

$a_3 = 6\cdot(-2)^{3-1} = 6\cdot(-2)^2 = 6\cdot4 = 24$

Step4: Calculate n=4 term

$a_4 = 6\cdot(-2)^{4-1} = 6\cdot(-2)^3 = 6\cdot(-8) = -48$

Answer:

6, -12, 24, -48