Answer
$\dfrac{2}{5},\dfrac{2}{3},\dfrac{6}{7},1$
Work Step by Step
Need to find the first four terms of $a_n=\dfrac{2n}{n+4}$
when $n=1,2,3,4$
Thus, $a_1=\dfrac{2(1)}{1+4}=\dfrac{2}{5}$
$a_2=\dfrac{2(2)}{2+4}=\dfrac{4}{6}=\dfrac{2}{3}$
$a_3=\dfrac{2(3)}{3+4}=\dfrac{6}{7}$
$a_4=\dfrac{2(4)}{4+4}=1$ Hence, the first four terms are: $\dfrac{2}{5},\dfrac{2}{3},\dfrac{6}{7},1$
