Given: $a_n=3(a_{n−1})−1$
This means $a_6=3(a_{6−1})−1 = 3(a_5) - 1$
Unfortunately, we don't know the value of $a_5$. So what do we do?

We say that the defined terms are recursive, which means we must work our way up to determining the value of $a_6$.

Given: $a_1=1$

$a_2=3(a_{2−1})−1=3(a_{1})−1 = 3(1) - 1 = 3 - 1 = 2$
$a_3=3(a_{3−1})−1=3(a_{2})−1 = 3(2) - 1 = 6 - 1 = 5$
$a_4=3(a_{3})−1 = 3(5) - 1 = 15 - 1 = 14$
$a_5=3(a_{4})−1 = 3(14) - 1 = 42 - 1 = 41$
$a_6=3(a_{5})−1 = 3(41) - 1 = 123 - 1 = 122$

Answer: D