Answer:
Given the sequence:
[tex]a_1 =2[/tex]
[tex]a_n = 3a_{n-1}-1[/tex]
where, n is the number of terms.
We have to find the first five terms of the sequence.
For n = 2;
[tex]a_2 = 3a_{2-1}-1[/tex]
⇒[tex]a_2 = 3a_{1}-1[/tex]
Substitute the given values we have;
[tex]a_2 = 3 \cdot 2 -1 = 6-1 = 5[/tex]
For n = 3
[tex]a_3 = 3a_{3-1}-1[/tex]
⇒[tex]a_3= 3a_{2}-1[/tex]
Substitute the given values we have;
[tex]a_3 = 3 \cdot 5 -1 = 15-1 = 14[/tex]
For n = 4
[tex]a_4 = 3a_{4-1}-1[/tex]
⇒[tex]a_4= 3a_{3}-1[/tex]
Substitute the given values we have;
[tex]a_4 = 3 \cdot 14 -1 = 42-1 = 41[/tex]
For n = 5
[tex]a_5 = 3a_{5-1}-1[/tex]
⇒[tex]a_5= 3a_{4}-1[/tex]
Substitute the given values we have;
[tex]a_5= 3 \cdot 41 -1 = 123-1 = 122[/tex]
Therefore, the first five terms of the sequence are:
[tex]a_1=2[/tex], [tex]a_2=5[/tex], [tex]a_3 =14[/tex], [tex]a_4 =41[/tex] and [tex]a_5=122[/tex],
