Answer
$a_{n}=a(1+i)^{n-1}$
Work Step by Step
(see p. 858)
A geometric sequence is a sequence whose terms are obtained by multiplying each term by the same fixed constant $r$ to get the next term.
A geometric sequence has the form $a, ar, ar^{2}, ar^{3}, \ldots$
The number $a$ is the first term of the sequence,
and the number $r $is the common ratio.
The nth term of the sequence is $\quad a_{n}=ar^{n-1}$
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$a_{1}=a=2$
$a_{2}=2(1+i)=2+2i$
$a_2=a(1+i)$
$a_{3}= a_{2}(1+i)=(2+2i)(1+i)=2+4i+2i^{2}=4i$
$a_{3}=a_{2}(1+i)=a(1+i)^{2}$,
$a_{4}= a_{3}(1+i)=4i(1+i)=4i+4i^{2}=-4+4i$
$a_{4}=a_{3}(1+i)=a(1+i)^{3}$,
...
$a_{n}=a(1+i)^{n-1}$
