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