Correct Answer – `-1`
We have,
`a_(n+1)=1/(1-a_(n))`
`thereforea_(2)=1/(1-a_(1))`
and `a_(3)=1/(1-a_(2))=1/(1-1/(1-a_(1))=(1-a_(1))/(-a_(1))`
Since `a_(3)=a_(1),` we have `1-a_(1)/(-a_(1))=a_(1)`
`rArra_(1)^(2)-a_(1)+1=0`
`rArra_(1)=-omega` or `-omega^(2)`, where `omega` is cube root of unity.
Now, `a_(5)=1/(1-a_(4))=1/(1-1/(1-a_(3)))`
`=(1-a_(3))/(-a_(3))`
`=(1-a_(1))/(-a_(1))=a_(1)=a_(3)` and so on
`thereforea_(1)=a_(3)=a_(5)….a_(2001)`
Thus, `(a_(2001))^(2001)=(-omega)^(2001)`
or `(-omega^(2))^(2001=-1`
or `(-1)^(2001)(omega^(3))^(1334)=-1` (`becauseomega^(3)=1`)

`a_2=1/(1-a_1)`
`a_3=1/(1-a_2)=(1-a_1)/(-a_1)`
`-a_1^2=(1-a_1)`
`a_1^2-a_1+1=0`
`a_1=(1pmsqrt(1-4))/2=e^(ipi/3),e^(i-pi/3`
`a_1=a_3=a_5=a_7…=a_2001=e^(ipi/3`
`(a_2001)^2001=(e^ipmpi/3)^2001`
`=e^(pmi(667pi)`
`=cos(667pi)pmsin(667pi)`
`=-1`.