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若 \({\left( {1 - 4x} \right)^{2021}} = {a_0} + {a_1}x + {a_2}{x^2} + \cdots + {a_{2021}}{x^{2021}}\)，则 \(\frac{{{a_1}}}{2} + \frac{{{a_2}}}{{{2^2}}} + \cdots + \frac{{{a_{2021}}}}{{{2^{2021}}}}\) 的值是（ ）
A.\( - 2\)
B.\( - 1\)
C.\(0\)
D.\(1\)
参考答案：A
解析：
令 \(x = 0\)，得 \({a_0} = 1\)，
令 \(x = \frac{1}{2}\)，得 \({a_0} + \frac{{{a_1}}}{2} + \frac{{{a_2}}}{{{2^2}}} + \cdots + \frac{{{a_{2021}}}}{{{2^{2021}}}} = {\left( { - 1} \right)^{2021}} = - 1\)，
则 \(\frac{{{a_1}}}{2} + \frac{{{a_2}}}{{{2^2}}} + \cdots + \frac{{{a_{2021}}}}{{{2^{2021}}}} = {\left( { - 1} \right)^{2021}} - {a_0} = - 1 - 1 = - 2\).
故选：A
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