Question:
Let < an > be a sequence defined by
a1 = 3
and, an = 3an − 1 + 2, for all n > 1
Find the first four terms of the sequence.
Solution:
Given:
a1 = 3
And, an = 3an − 1 + 2 for all n > 1
$a_{2}=3 a_{2-1}+2=3 a_{1}+2=11$
$a_{3}=3 a_{3-1}+2=3 a_{2}+2=35$
$a_{4}=3 a_{4-1}+2=3 a_{3}+2=107$
Thus, the first four terms of the sequence are $3,11,35,107$.
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