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