Tính tổng cấp số nhân lùi vô hạn\[{\rm{S = 1 + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{ }}{{\rm{2}}^{\rm{2}}}}}{\rm{ + }}...{\rm{ + }}\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{n}}}}}{\rm{ + }}...\]

A. q = 2

B. q = −4

C. q = 4

A. \(\left\{ {\begin{array}{*{20}{c}}{q\,\,{\rm{ = 2}}}\\{{{\rm{u}}_1}{\rm{ = 4}}}\end{array}} \right.\)

B. \(\left\{ {\begin{array}{*{20}{c}}{q\,\,{\rm{ = 2}}}\\{{{\rm{u}}_1}{\rm{ = 6}}}\end{array}} \right.\)

C. \(\left\{ {\begin{array}{*{20}{c}}{q\,\,{\rm{ = 2}}}\\{{{\rm{u}}_1}{\rm{ = 9}}}\end{array}} \right.\)

D. \(\left\{ {\begin{array}{*{20}{c}}{q\,\,{\rm{ = 3}}}\\{{{\rm{u}}_1}{\rm{ = 9}}}\end{array}} \right.\)

A. \[{{\rm{u}}_{\rm{n}}}{\rm{ = }} - \frac{{\rm{1}}}{{{{\rm{2}}^{\rm{n}}}}}\]

B. \[{{\rm{u}}_{\rm{n}}}{\rm{ = }}{{\rm{2}}^{{\rm{n}} - {\rm{2}}}}\]

C. \[{{\rm{u}}_{\rm{n}}}{\rm{ = }} - {{\rm{2}}^{{\rm{n}} - {\rm{1}}}}\]

D. \[{{\rm{u}}_{\rm{n}}}{\rm{ = }} - \frac{{\rm{1}}}{{{{\rm{2}}^{{\rm{n}} - {\rm{1}}}}}}\]

A. 11

B. 12

C. 10

D. 13

A. \[{{\rm{u}}_{\rm{n}}}{\rm{ = 7}} - {\rm{3n}}\]

B. \[{{\rm{u}}_{\rm{n}}}{\rm{ = 7}} - {{\rm{3}}^{\rm{n}}}\]

C. \[{{\rm{u}}_{\rm{n}}}{\rm{ = }}\frac{{\rm{7}}}{{{\rm{3n}}}}\]

D. \[{{\rm{u}}_{\rm{n}}}{\rm{ = 7}}{\rm{.}}{{\rm{3}}^{\rm{n}}}\]

A. 6 m²

B. 8 m²

C. 10 m²

D. 12 m²

A. \[\frac{{{{10}^{\rm{n}}} - 1}}{{81}}\]

B. \[10\left( {\frac{{{{10}^{\rm{n}}} - 1}}{{81}}} \right)\]

C. \[\left[ {10\left( {\frac{{{{10}^{\rm{n}}} - 1}}{{81}}} \right) - {\rm{n}}} \right]\]

D. \[\frac{1}{9}\left[ {10\left( {\frac{{{{10}^{\rm{n}}} - 1}}{9}} \right) - {\rm{n}}} \right]\]