若复数a+3i1+2i（a∈R，i为虚数单位）是纯虚数，则limn→∞(1a+1a2+…+1an)=（）A．17B．57C．-1

Question

若复数 (a+3i)/(1+2i) （a∈R，i为虚数单位）是纯虚数，则

lim

n→∞

((1)/(a) +(1)/(a2) +…+ (1)/(an) ) =（　　）

A． (1)/(7)

B． (5)/(7)

C．- (1)/(7)

D．- (5)/(7)

Answer 1

(a+3i)/(1+2i) = ((a+3i)(1-2i))/(5)
= (a+3i-2ai+6)/(5)
= (a+6)/(5) +(3-2a)/(5) i ，
∵复数 (a+3i)/(1+2i) = (a+6)/(5) +(3-2a)/(5) i 是纯虚数，
∴

(a+6)/(5) =0

(3-2a)/(5) ≠ 0

，
解得a=-6．
∴

lim

n→∞

((1)/(a) +(1)/(a2) +…+ (1)/(an) )
=

lim

n→∞

(-(1)/(6)[1- (-(1)/(6))n ])/(1-(-(1)/(6)))
=- (1)/(7) ．
故选C．