Đáp án:

 $2$

Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
G = \left( {1 + \dfrac{1}{2}} \right)\left( {1 + \dfrac{1}{4}} \right)\left( {1 + \dfrac{1}{{16}}} \right)...\left( {1 + \dfrac{1}{{{2^{1024}}}}} \right)\\
 \Rightarrow G = \left( {1 + \dfrac{1}{2}} \right)\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^2}} \right)\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^4}} \right)...\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^{{2^{10}}}}} \right)\\
 \Rightarrow \left( {1 - \dfrac{1}{2}} \right)G = \left( {1 - \dfrac{1}{2}} \right)\left( {1 + \dfrac{1}{2}} \right)\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^2}} \right)\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^4}} \right)...\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^{{2^{10}}}}} \right)
\end{array}$

Áp dụng hằng đẳng thức ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right),\forall a,b$

Khi đó:

$\begin{array}{l}
\left( {1 - \dfrac{1}{2}} \right)G = \left( {1 - {{\left( {\dfrac{1}{2}} \right)}^2}} \right)\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^2}} \right)\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^4}} \right)...\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^{{2^{10}}}}} \right)\\
 \Rightarrow \dfrac{1}{2}G = \left( {1 - {{\left( {\dfrac{1}{2}} \right)}^4}} \right)\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^4}} \right)...\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^{{2^{10}}}}} \right)\\
 = \left( {1 - {{\left( {\dfrac{1}{2}} \right)}^{{2^{10}}}}} \right)\left( {1 + {{\left( {\dfrac{1}{2}} \right)}^{{2^{10}}}}} \right)\\
 = 1 - {\left( {\dfrac{1}{2}} \right)^{{2^{11}}}}\\
 = 1 - \dfrac{1}{{{2^{{2^{11}}}}}}\\
 = 1 - \dfrac{1}{{{2^{2048}}}}\\
 \Rightarrow \dfrac{1}{2}G = 1 - \dfrac{1}{{{2^{2048}}}} \Rightarrow G = 2 - \dfrac{2}{{{2^{2048}}}} = 2 - \dfrac{1}{{{2^{2047}}}}\\
 \Rightarrow G = 2 - H\left( {Do:H = \dfrac{1}{{{2^{2047}}}}} \right)\\
 \Rightarrow G + H = 2
\end{array}$

Vậy $G+H=2$