Question

复数数列{an} 满足a1=0 ,an=[a(下标）n-1]^2+i (n>=2,i 为虚数单位),则它的前2007 项

Answer 1

a1=0

a2=a1^2+i=i

a3=i^2+i=i-1

a4=(i-1)^2+i=-i

a5=(-i)^2+i=i-1

a6=(i-1)^2+i=-i

.

a1+a2+a3+...+a2007=0+i+i-1-i+i+1-i.+i-1

=i-1002+i-1

=2i-1003