By iterative method:
$T(n)=2T(n-1)-1 =4T(n-2)-3 =8T(n-3)-7 ....=2^iT(n-i)-(2^i-1)$
The algorithm converges when T(n)=T(0), i.e when i=n.
when i=n, $T(n)=2^iT(n-i)-(2^i-1)$ become
$T(n)=2^nT(0)-(2^n-1)$
=$2^n-(2^i-1)$
=$2^n-2^i+1$
=1
i.e O(1).
Answer is D none of the above option.