University Calculus: Early Transcendentals (3rd Edition)

$a_1=1$ $a_2=3$ $a_3=1$ $a_4=3$
We know that $a_n=2+(-1)^n$. Hence, here: $a_1=2+(-1)^1=2+(-1)=1$ $a_2=2+(-1)^2=2+1=3$ $a_3=2+(-1)^3=2+(-1)=1$ $a_4=2+(-1)^4=2+1=3$