Gọi q là công bội của cấp số nhân \(\left( {{{\rm{u}}_{\rm{n}}}} \right)\). Ta có \({{\rm{S}}_2} = 4 \ne 0\) nên \({{\rm{u}}_1} \ne 0\).

Ta có \({{\rm{u}}_2} < 0 \Leftrightarrow {{\rm{u}}_1}{\rm{q}} < 0 \Leftrightarrow {{\rm{u}}_1}\) và \({\rm{q}}\)trái dấu.

Ta có:\(\left\{ {\begin{array}{*{20}{l}}{{{\rm{S}}_2} = 4}\\{\;{{\rm{S}}_3} = 13}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{{\rm{u}}_1} + {{\rm{u}}_2} = 4}\\{{{\rm{u}}_1} + {{\rm{u}}_2} + {{\rm{u}}_3} = 13}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{u_1} + {u_1}q = 4}\\{{u_1} + {u_1}q + {u_1}{q^2} = 13}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{u_1} + {u_1}q = 4}\\{13{u_1}(1 + q) = 4{u_1}\left( {1 + q + {q^2}} \right)}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{u_1} + {u_1}q = 4}\\{4{q^2} - 9q - 9 = 0}\end{array} \Leftrightarrow \left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{{u_1} = 1}\\{q = 3}\end{array}\,\,\,({\rm{ktm}})} \right.\\\left\{ \begin{array}{l}{u_1} = 16\\q = - \frac{3}{4}\,\,(tm)\end{array} \right.\end{array} \right.} \right.\).

Vậy \({{\rm{S}}_6} = \frac{{{{\rm{u}}_1}\left( {1 - {{\rm{q}}^6}} \right)}}{{1 - {\rm{q}}}} = \frac{{16\left( {1 - {{\left( { - \frac{3}{4}} \right)}^6}} \right)}}{{1 + \frac{3}{4}}} = \frac{{481}}{{64}}{\rm{. }}\)Chọn A.