Intermediate Algebra (6th Edition)

$0, -1, 2, -3, 4$
Let's evaluate $a_1, a_2, a_3, a_4,$and $a_5$ by plugging in $1, 2, 3, 4,$ and $5$ for each $n$. We use PEMDAS (parentheses, exponentation, multiplication, division, addition, subtraction) to decide which operations go first. \begin{align} a_1 &= (-1)^{1+1}(1-1) = (-1)^2(1-1) = 1 \cdot(1-1) = 1 \cdot 0 = 0\\ \\ a_2 &= (-1)^{2+1}(2-1) = (-1)^3(2-1) = -1\cdot(2-1) = -1 \cdot 1 = -1\\ \\ a_3 &= (-1)^{3+1}(3-1) = (-1)^4(3-1) = 1 \cdot(3-1) = 1 \cdot 2 = 2\\ \\ a_4 &= (-1)^{4+1}(4-1) = (-1)^5(4-1) = -1\cdot(4-1) = -1 \cdot 3 = -3\\ \\ a_5 &= (-1)^{5+1}(5-1) = (-1)^6(5-1) = 1 \cdot(5-1)= 1 \cdot 4 = 4\\ \\ \end{align}