B=[tex](1+\frac{1}{2})(1+\frac{1}{2^{2}})(1+\frac{1}{2^{2.2}})...(1+\frac{1}{2n})[/tex]
$(1-\frac{1}{2})B=(1-\frac{1}{2})(1+\frac{1}{2})(1+\frac{1}{2^{2}})(1+\frac{1}{2^{2.2}})...(1+\frac{1}{2^{2n}}) = (1-\frac{1}{2^{2}})(1+\frac{1}{2^{2}})(1+\frac{1}{2^{2.2}})...(1+\frac{1}{(2^{n})^{2}})=1-\frac{1}{(2^{n})^{4}}=1-\frac{1}{2^{4n}}$
=> $B= 2-\frac{1}{2^{4n-1}}$