Answer
The first five terms are:
$a_1 = -2
\\a_2 = 4
\\a_3 = -8
\\a_4= 16
\\a_5= -32$
The sequence is geometric with $r=-2$.
The $n^{th}$ term is: $a_n = -2 \cdot(-2)^{n-1}$
Work Step by Step
To find the first five terms, substitute 1, 2, 3, 4, and 5 to the given formula to obtain:
$a_1 = (-1)^1(2^1) = -1(2) = -2
\\a_2 = (-1)^2(2^2) = 1(4) = 4
\\a_3 = (-1)^3(2^3) = -1(8) = -8
\\a_4=(-1)^4(2^4) = 1(16) = 16
\\a_5=(-1)^5(2^5) = -1(32)=-32$
RECALL:
A sequence is geometric if there is a common ratio among consecutive terms.
Solve for the ratio of each pair of consecutive terms to obtain:
$\dfrac{4}{-2} = -2
\\\dfrac{-8}{4} = -2
\\\dfrac{16}{-8} = -2
\\\dfrac{-32}{16}=-2$
The ratio is common therefore the sequence is geometric with $r=-2$.
The $n^{th}$ term of a geometric sequence is given by the formula $a_n = ar^{n-1}$
where $a$ = first term and $r$= common ratio.
This means that the $n^{th}$ term of the given geometric sequence, whose $a=-2$ and $r=-2$, is:
$a_n = -2 \cdot(-2)^{n-1}$
