## Algebra 2 (1st Edition)

n-th term:$\quad a_{n}=(-1)^{n}\cdot 4n$ The next term is $\quad a_{5}=-20$
The sign alternates, $(-,+,-,+,-,....)$ We use the fact that the signs of powers of $(-1)$ also alternate. $\left[\begin{array}{cccccccc} n: & 1 & 2 & 3 & 4\\ a_{n}: & (-1)^{1}\cdot 4 & (-1)^{2}\cdot 8, & (-1)^{3}\cdot 12, & (-1)^{4}\cdot 16 \end{array}\right]$ $\Rightarrow (-1)^{n}$ is a factor of the nth term (The nth term is negative for odd n's and positive for even n's) The absolute values of the terms are multiples of 4, $\left[\begin{array}{cccccccc} n: & 1 & 2 & 3 & 4 & ...\\ a_{n}: & (-1)^{1}\cdot 4\cdot 1 & (-1)^{2}\cdot 4\cdot 2, & (-1)^{3}\cdot 4\cdot 3, & (-1)^{4}\cdot 4\cdot 4 & \end{array}\right]$ $\Rightarrow (4n)$ is a factor of the nth term The n-th term is: $a_{n}=(-1)^{n}\cdot 4n$ The next term is $a_{5}=-4(5)=-20$