u n − 1 = 1042 .

Ta có \({u_{n - 1}} = \frac{{{u_n}}}{q} = \frac{{2048}}{2} = 1024\). Vì \({u_n} = {u_1} \cdot {q^{n - 1}} \Leftrightarrow 2048 = {u_1} \cdot {2^{n - 1}} \Leftrightarrow 4096 = {u_1} \cdot {2^n}\).

Ta có \({u_1} + {u_2} + {u_3} + .... + {u_{n - 1}} + {u_n} = {S_n} = 4092\).

\( \Rightarrow {u_1} + {u_2} + {u_3} + ... + {u_{n - 1}} = 4092 - {u_n} = 4092 - 2048 = 2044\).

Ta có \(\left\{ \begin{array}{l}{u_1} \cdot \frac{{1 - {q^n}}}{{1 - q}} = 4092\\{u_1} \cdot {q^{n - 1}} = 2048\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1}\left( {1 - {q^n}} \right) = 4092\left( {1 - q} \right)\\{u_1} \cdot {q^n} = 2048q\end{array} \right.\).

Từ đó suy ra \({u_1} - 2048q = 4092 - 4092q \Leftrightarrow {u_1} = - 2044q + 4092 = - 2044 \cdot 2 + 4092 = 4\).

Khi đó \({u_7} = {u_1} \cdot {q^6} = 4 \cdot {2^6} = 256\).

Đáp án:           a) Sai,            b) Đúng,         c) Đúng,          d) Sai.