The answer is 2390.

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write as: $\dfrac{1}{2}\left(\dfrac{2^n}{1+2^n+2^{n+1}(1+2^n)}\right)$ $=\dfrac{1}{2}\left(\dfrac{2^{n+1}+1-(2^n+1)}{(1+2^n)(1+2^{n+1})}\right)$ $=\dfrac{1}{2}\left(\dfrac{1}{2^n+1}-\dfrac{1}{2^{n+1}+1}\right)$ so the summation becomes: $\sum_{n=1}^{10} \left(\dfrac{1}{2}\left(\dfrac{1}{2^n+1}-\dfrac{1}{2^{n+1}+1}\right)\right)$ this easily telescopes to $\dfrac{1}{2}\left(\dfrac{1}{2^1+1}-\dfrac{1}{2^{11}+1}\right)=\dfrac{341}{2049}$ $341+2049=\boxed{2390}$