Recurrence Relation:\[\] The given recurrence relation is: \[\] \( a_n = 2a_{n-1} + 3 \) \[\] with the initial condition: \( a_0 = 6 \) \[\] We will compute \( a_1, a_2, a_3, a_4 \) step by step:\[\] \[\] Step 1: Compute ( a_1 ) \[\] \(a_1 = 2a_0 + 3 = 2(6) + 3 = 12 + 3 = 15\) \[\] \[\] Step 2: Compute \( a_2 \) \[\] \( a_2 = 2a_1 + 3 = 2(15) + 3 = 30 + 3 = 33\) \[\] \[\] Step 3: Compute \( a_3 \) \[\] \(a_3 = 2a_2 + 3 = 2(33) + 3 = 66 + 3 = 69\) \[\] \[\] Step 4: Compute \( a_4 \) \[\] \(a_4 = 2a_3 + 3 = 2(69) + 3 = 138 + 3 = 141\) \[\] \[\] Final Answer:\[\] \[\] Correct Option: (c)