(1)∵z1=a1+b1•i=1+i,∴a1=1,b1=1. 由zn+1=2zn++2i,得an+1+bn+1•i=2(an+bn•i)+(an-bn•i)+2i=3an+(bn+2)•i, ∴, ∴数列{an}是以1为首项公比为3的等比数列,数列{bn}是以1为首项公差为2的等差数列, ∴an=3n-1,bn=2n-1; (2)由(1)知an=3n-1,bn=2n-1. ①z1+z2+…+zn=(a1+a2+…+an)+(b1+b2+…+bn)•i =(1+31+32+…+3n-1)+(1+3+5+••+2n-1)•i =(3n-1)+n2•i. ②令Sn=a1b1+a2b2+…+anbn,Sn=1+3•3+32•5+…+3n-1•(2n-1)(Ⅰ) 将(Ⅰ)式两边乘以3得,3Sn=3•1+32•3+33•5+…+3n•(2n-1)(Ⅱ) 将(Ⅰ)减(Ⅱ)得-2Sn=1+2•3+2•32+2•33+…+2•3n-1-3n•(2n-1). ∴-2Sn=-2+3n(-2n+2), 所以Sn=(n-1)•3n+1. |