jmc

First, we can factor the denominator: $$1+2^n+2^{n+1}+2^{2n+1}=(1+2^n)+2^{n+1}(1+2^n)=(1+2^n)(1+2^{n+1}).$$Then we can write the numerator $2^n$ as $(1+2^{n+1})-(1+2^n)=2^n,$ so $$\frac{2^n}{1+2^n+2^{n+1}+2^{2n+1}}=\frac{(1+2^{n+1})-(1+2^n)}{(1+2^n)(1+2^{n+1})}=\frac{1}{1+2^n}-\frac{1}{1+2^{n+1}}.$$Hence, $$\begin{array}{rl}\sum _{n=1}^{\infty }\frac{2^n}{1+2^n+2^{n+1}+2^{2n+1}}& =\left(\frac{1}{1+2}-\frac{1}{1+2^2}\right)+\left(\frac{1}{1+2^2}-\frac{1}{1+2^3}\right)+\left(\frac{1}{1+2^3}-\frac{1}{1+2^4}\right)+\cdots \\ & =\boxed{\frac{1}{3}}.\end{array}$$