Dãy số (u_n) với u_n = $\frac{1}{\mathrm{n}}-2$, là 

Trong \[\left( {ABCD} \right)\], gọi \[E = AC \cap BM\]

Ta có: \[\left\{ \begin{array}{l}E \in AC,AC \subset \left( {SAC} \right)\\E \in BM,BM \subset \left( {SBM} \right)\end{array} \right. \Rightarrow E \in \left( {SAC} \right) \cap \left( {SBM} \right){\rm{ }}\left( 1 \right)\]

Ta có:  \[\left\{ \begin{array}{l}S \in \left( {SAC} \right)\\S \in \left( {SBM} \right)\end{array} \right. \Rightarrow S \in \left( {SAC} \right) \cap \left( {SBM} \right){\rm{ }}\left( 2 \right)\]

Từ \[\left( 1 \right)\] và \[\left( 2 \right)\], ta có: \[SE = \left( {SAC} \right) \cap \left( {SBM} \right)\]

Chọn D

Ta có: \(\left\{ \begin{array}{l}{u_9} = 5{u_2}\\{u_{13}} = 2{u_6} + 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1} + 8d = 5({u_1} + d)\\{u_1} + 12d = 2({u_1} + 5d) + 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 4{u_1} + 3d = 0\\ - {u_1} + 2d = 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 3\\d = 4\end{array} \right.\)

\({S_{11}} = \frac{{11}}{2}(2.3 + 10.4) = 253\).、