Suppose that F(n+1)=2F(n)+12F(n+1)=\dfrac{2F(n)+1}{2}F(n+1)=22F(n)+1 for n=1,2,3,…n=1,2,3,\ldotsn=1,2,3,… and F(1)=2F(1)=2F(1)=2. Then F(101)F(101)F(101) equals A505050B525252C545454D565656E585858FNone of the above