a) Ta có \({v_n} = {u_{n + 1}} - {u_n} = \left( {{u_n} + \frac{1}{{{2^n}}}} \right) - {u_n} = \frac{1}{{{2^n}}}\)

Khi đó \(A = {v_1} + {v_2} + \ldots + {v_n} = \frac{1}{2} + \frac{1}{{{2^2}}} + \ldots + \frac{1}{{{2^n}}}\)

\( \Rightarrow \frac{A}{2} = \frac{1}{2}\left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \ldots + \frac{1}{{{2^n}}}} \right)\)

\( \Rightarrow A - \frac{A}{2} = \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \ldots + \frac{1}{{{2^n}}}} \right) - \left( {\frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \ldots + \frac{1}{{{2^{n + 1}}}}} \right) = \frac{1}{2} - \frac{1}{{{2^{n + 1}}}}\)

\( \Rightarrow A = 1 - \frac{1}{{{2^n}}}\).

b) Từ câu a, suy ra \(A = {v_1} + {v_2} + \ldots + {v_n} = {u_2} - {u_1} + {u_3} - {u_2} + \ldots + {u_n} - {u_{n - 1}} + {u_{n + 1}} - {u_n}\)

\( \Leftrightarrow A = \sum\limits_{i = 1}^n {{v_i}} = {u_{n + 1}} - {u_1} = {u_{n + 1}} - 1 = {u_n} + \frac{1}{{{2^n}}} - 1 \Rightarrow 1 - \frac{1}{{{2^n}}} = {u_n} + \frac{1}{{{2^n}}} - 1\)

\( \Rightarrow {u_n} = 2 - \frac{1}{{{2^{n - 1}}}}\).

c) \(\lim {u_n} = \lim \left( {2 - \frac{1}{{{2^{n - 1}}}}} \right) = 2\).

Do đó ta chọn đáp án như sau