Series: (2+(1-i)+\dfrac{1}{i}+\cdots) to 8 terms.

First check if G.P.: [ r_1=\dfrac{1-i}{2},\quad r_2=\dfrac{1/i}{1-i} ] (\dfrac{1}{i}=-i), so second ratio (\dfrac{-i}{1-i}).

These may not be equal. Interpreting as G.P. with first term (a=2) and finding (r) from consecutive terms carefully.

If terms are (2,\ 1-i,\ -i,\ \ldots) and common ratio from (a_2/a_1): [ r=\dfrac{1-i}{2} ] Then (a_3=2\cdot\left(\dfrac{1-i}{2}\right)^2=\dfrac{(1-i)^2}{2}=\dfrac{1-2i-1}{2}=-i=\dfrac{1}{i}) ✓

So G.P. with (a=2), (r=\dfrac{1-i}{2})

[ S_8=2\cdot\dfrac{1-r^8}{1-r} ]

Answer: (S_8=2\dfrac{1-\left(\dfrac{1-i}{2}\right)^8}{1-\dfrac{1-i}{2}})