Hệ số A1 bằng \(1 - \frac{1}{{{2^{9999}}}}\) .

Hệ số A2 bằng \[\frac{{{4^{9999}} - {{3.2}^{9999}} + 2}}{{{{3.4}^{9999}}}}\] .

Giải thích

Hệ số của \[{x^{9998}}\] là: \[{A_1} = \frac{1}{2} + \frac{1}{{{2^2}}} +  \ldots  + \frac{1}{{{2^{9999}}}} = \frac{1}{2}.\frac \right)}^{9999}}}}} = 1 - \frac{1}{{{2^{9999}}}}\] . 

Hệ số của \[{x^{9997}}\] là \[{A_2} = \frac{1}{2}.\frac{1}{{{2^2}}} + \frac{1}{{{2^2}}}.\frac{1}{{{2^3}}} +  \ldots  + \frac{1}{{{2^{9998}}}}.\frac{1}{{{2^{9999}}}}\]

Ta có: \[A_1^2 = {\left( {\frac{1}{2} + \frac{1}{{{2^2}}} +  \ldots  + \frac{1}{{{2^{9999}}}}} \right)^2}\]

\[ \Leftrightarrow {\left( {1 - \frac{1}{{{2^{9999}}}}} \right)^2} = \frac{1}{4} + \frac{1}{{{4^2}}} +  \ldots  + \frac{1}{{{4^{9999}}}} + 2.\frac{1}{2}.\frac{1}{{{2^2}}} + 2.\frac{1}{{{2^2}}}.\frac{1}{{{2^3}}} +  \ldots  + 2.\frac{1}{{{2^{9998}}}}.\frac{1}{{{2^{9999}}}}\]

\[ \Leftrightarrow 1 - 2.1.\frac{1}{{{2^{9999}}}} + \frac{1}{{{4^{9999}}}} = \frac{1}{4}.\frac \right)}^{9999}}}}} + 2B\]

\[ \Leftrightarrow 1 - 2.1.\frac{1}{{{2^{9999}}}} + \frac{1}{{{4^{9999}}}} = \frac{1}{4}.\frac \right)}^{9999}}}}} + 2B\]

\[ \Leftrightarrow 1 - \frac{1}{{{2^{9998}}}} + \frac{1}{{{4^{9999}}}} = \frac{1}{3}.\left( {1 - \frac{1}{{{4^{9999}}}}} \right) + 2B\]

\[ \Leftrightarrow B = \frac{{{4^{9999}} - {{3.2}^{9999}} + 2}}{{{{3.4}^{9999}}}}\]