Đáp án đúng: B
Giải chi tiết:\(A = \frac{1}{{{5^2}}} + \frac{2}{{{5^3}}} + \frac{3}{{{5^4}}} + \ldots + \frac{n}{{{5^{n + 1}}}} + \ldots + \frac{{11}}{{{5^{12}}}}\)
\( \Rightarrow 5A = 5 \cdot \left( {\frac{1}{{{5^2}}} + \frac{2}{{{5^3}}} + \frac{3}{{{5^4}}} + \ldots + \frac{n}{{{5^{n + 1}}}} + \ldots + \frac{{11}}{{{5^{12}}}}} \right)\)\( = \frac{1}{5} + \frac{2}{{{5^2}}} + \frac{3}{{{5^3}}} + \ldots + \frac{n}{{{5^n}}} + \ldots + \frac{{11}}{{{5^{11}}}}\)
\(\begin{array}{l} \Rightarrow 5A = \frac{1}{5} + \frac{2}{{{5^2}}} + \frac{3}{{{5^3}}} + \ldots + \frac{n}{{{5^n}}} + \ldots + \frac{{11}}{{{5^{11}}}}\\\,\,\,\,\,\,\,\,A = \frac{1}{{{5^2}}} + \frac{2}{{{5^3}}} + \frac{3}{{{5^4}}} + \ldots + \frac{n}{{{5^{n + 1}}}} + \ldots + \frac{{11}}{{{5^{12}}}}\\ \Rightarrow 5A - A = \left( {\frac{1}{5} + \frac{2}{{{5^2}}} + \frac{3}{{{5^3}}} + \ldots + \frac{n}{{{5^n}}} + \ldots + \frac{{11}}{{{5^{11}}}}} \right) - \left( {\frac{1}{{{5^2}}} + \frac{2}{{{5^3}}} + \frac{3}{{{5^4}}} + \ldots + \frac{n}{{{5^{n + 1}}}} + \ldots + \frac{{11}}{{{5^{12}}}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{5} + \frac{2}{{{5^2}}} + \frac{3}{{{5^3}}} + \ldots + \frac{n}{{{5^n}}} + \ldots + \frac{{11}}{{{5^{11}}}} - \frac{1}{{{5^2}}} - \frac{2}{{{5^3}}} - \frac{3}{{{5^4}}} - \ldots - \frac{n}{{{5^{n + 1}}}} - \ldots - \frac{{11}}{{{5^{12}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{5} + \left( {\frac{2}{{{5^2}}} - \frac{1}{{{5^2}}}} \right) + \left( {\frac{3}{{{5^3}}} - \frac{2}{{{5^3}}}} \right) + \ldots + \left( {\frac{n}{{{5^n}}} - \frac{{n - 1}}{{{5^n}}}} \right) + \ldots + \left( {\frac{{11}}{{{5^{11}}}} - \frac{{10}}{{{5^{11}}}}} \right) - \frac{{11}}{{{5^{12}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + \ldots + \frac{1}{{{5^n}}} + \ldots + \frac{1}{{{5^{11}}}} - \frac{{11}}{{{5^{12}}}}\end{array}\)
\( \Rightarrow 4A = \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + \ldots + \frac{1}{{{5^n}}} + \ldots + \frac{1}{{{5^{11}}}} - \frac{{11}}{{{5^{12}}}}\)
\( \Rightarrow 4A = \left( {\frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + \ldots + \frac{1}{{{5^n}}} + \ldots + \frac{1}{{{5^{11}}}}} \right) - \frac{{11}}{{{5^{12}}}}\)
Đặt \(B = \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + \ldots + \frac{1}{{{5^n}}} + \ldots + \frac{1}{{{5^{11}}}}\).
Biểu thức \(A\) trở thành: \(4A = B - \frac{{11}}{{{5^{12}}}}\)
Tính tổng: \(B = \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + \ldots + \frac{1}{{{5^n}}} + \ldots + \frac{1}{{{5^{11}}}}\)
\( \Rightarrow 5B = 5 \cdot \left( {\frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + \ldots + \frac{1}{{{5^n}}} + \ldots + \frac{1}{{{5^{11}}}}} \right)\)
\( \Rightarrow 5B = 1 + \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + \ldots + \frac{1}{{{5^n}}} + \ldots + \frac{1}{{{5^{10}}}}\)
\(5B - B = \left( {1 + \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + \ldots + \frac{1}{{{5^n}}} + \ldots + \frac{1}{{{5^{10}}}}} \right) - \left( {\frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + \ldots + \frac{1}{{{5^n}}} + \ldots + \frac{1}{{{5^{11}}}}} \right)\)
\(\begin{array}{l} \Rightarrow 4B = 1 + \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + \ldots + \frac{1}{{{5^n}}} + \ldots + \frac{1}{{{5^{10}}}} - \frac{1}{5} - \frac{1}{{{5^2}}} - \frac{1}{{{5^3}}} - \ldots - \frac{1}{{{5^n}}} - \ldots - \frac{1}{{{5^{11}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \frac{1}{{{5^{11}}}} = \frac{{{5^{11}} - 1}}{{{5^{11}}}}\end{array}\)
\( \Rightarrow B = \frac{{{5^{11}} - 1}}{{{{4.5}^{11}}}}\)
Thay \(B = \frac{{{5^{11}} - 1}}{{{{4.5}^{11}}}}\) vào biểu thức \(A\) ta được:
\(\begin{array}{l}4A = \frac{{{5^{11}} - 1}}{{{{4.5}^{11}}}} - \frac{{11}}{{{5^{12}}}} = \frac{{5\left( {{5^{11}} - 1} \right)}}{{{{4.5}^{12}}}} - \frac{{44}}{{{{4.5}^{12}}}}\\\,\,\,\,\,\,\,\, = \frac{{{5^{12}} - 5 - 44}}{{{{4.5}^{12}}}} = \frac{{{5^{12}} - 49}}{{{{4.5}^{12}}}}.\end{array}\)
\( \Rightarrow A = \frac{{{5^{12}} - 49}}{{{{16.5}^{12}}}} = \frac{1}{{16}} \cdot \frac{{{5^{12}} - 49}}{{{5^{12}}}}\)\( = \frac{1}{{16}} \cdot \left( {\frac{{{5^{12}}}}{{{5^{12}}}} - \frac{{49}}{{{5^{12}}}}} \right) = \frac{1}{{16}} \cdot \left( {1 - \frac{{49}}{{{5^{12}}}}} \right)\)
Vì \(1 - \frac{{49}}{{{5^{12}}}} < 1 \Rightarrow \frac{1}{{16}} \cdot \left( {1 - \frac{{49}}{{{5^{12}}}}} \right) < \frac{1}{{16}} \Rightarrow A < \frac{1}{{16}}\)
Vậy \(A < \frac{1}{{16}} \cdot \)
Chọn B.