Answer
$a_1=-\dfrac{4}{3} \\a_2=\dfrac{5}{4} \\a_3=-\dfrac{6}{5} \\a_4=\dfrac{7}{6} \\a_5=-\dfrac{8}{7}$
Work Step by Step
We are given:
$\{a_n\}=(-1)^n\dfrac{n+3}{n+2}$
In order to determine the remaining values, we will have to substitute $n=1, 2,3,4,5$ into the given sequence. Thus:
$a_1=(-1)^1\dfrac{1+3}{1+2}=-\dfrac{4}{3} \\a_2=(-1)^2\dfrac{2+3}{2+2}=\dfrac{5}{4} \\a_3=(-1)^3\dfrac{3+3}{3+2}=-\dfrac{6}{5} \\a_4=(-1)^4\dfrac{4+3}{4+2}=\dfrac{7}{6} \\a_5=(-1)^5\dfrac{5+3}{5+2}=-\dfrac{8}{7}$
