这是等比数列前n项之和，首项为A(1十i)^0，公比为(1十i)F＝A(1+i)^0+A(1+i)^1+A(1+i)^2+．．．+A(1+i)^n-2+A(1+i)^n-1 =A((1+i)^0+A(1+i)^1+A(1+i)^2+．．．+A(1+i)^(n-2)+A(1+i)^(n-1))=A((1+i)^n一1)/((1+i)一1)=A*{[(1+i)^n-1]/i}
owth has slowe
