Sequence recurrence relation solution

Recurrence: $a_{n+1} = 3a_n + 4n - 2$. Given $a_1 = 1$. $a_2 = 3(1) + 4(1) - 2 = 5$. $a_3 = 3(5) + 4(2) - 2 = 21$. Testing option C ($3^{n} - 2n$ or $3^n - 200$ for $n=100$): For $n=1$: $3^1 - 2(1) = 1 = a_1$. For $n=2$: $3^2 - 2(2) = 5 = a_2$. For $n=3$: $3^3 - 2(3) = 21 = a_3$. In general, $a_n = 3^n - 2n$. Therefore, $a_{100} = 3^{100} - 200$.