Question
Find the first four terms of the sequence defined by $\text{a}_1=3$ and $\text{a}_\text{n}=3\text{a}_{\text{n}-1}+2$ for all $\text{n}>1.$

Answer

$\text{a}_\text{n}=3\text{a}_{\text{n}-1}+2$ $​​\text{a}_1=3$ $​​\text{a}_1=​​\text{3a}_{2-1}+2=​\text{3a}_{2-1}+2=3(3)+2=11$ $​​\text{a}_3=​​\text{3a}_{3-1}+2=3​​\text{a}_2+2=3(11)+2=35$ $​​\text{a}_4=\text{3a}_{4-1}+2=\text{3a}_3+2=3(35)+2=107$ First four terms of the sequence are 3, 11, 35 and 107.

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