（2010•湖北模拟）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．-17D．-57

Question

（2010•湖北模拟）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．