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

A. \[\frac{{\sqrt 3 }}{2}\]

B. \[ - \frac{{\sqrt 3 }}{2}\]

C. \[\frac{1}{2}\]

D. \[ - \frac{1}{2}\]

A. \[\frac{{{{10}^{\rm{n}}} - 1}}{9}\]

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

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

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

A. m = 3

B. m = −4

C. m = 1

D. m = −3

A. 6 m²

B. 8 m²

C. 10 m²

D. 12 m²

A. −2; 10; 50; −250

B. −2; 10; −50; 250

C. −2; −10; −50; −250

D. −2; 10; 50; 250

A. \[{{\rm{u}}_{\rm{k}}}{\rm{ = }}{{\rm{u}}_{\rm{1}}}{\rm{ + }}\left( {{\rm{k}} - {\rm{1}}} \right){\rm{q}}\]

B. \[{{\rm{u}}_{\rm{k}}}{\rm{ = }}{{\rm{u}}_{\rm{1}}}{\rm{ + }}\left( {{\rm{k}} - {\rm{1}}} \right){\rm{q}}\]

C. \[{{\rm{u}}_{\rm{k}}}{\rm{ = }}\frac{{{{\rm{u}}_{{\rm{k}} - {\rm{1}}}}{\rm{ + }}{{\rm{u}}_{{\rm{k + 1}}}}}}{{\rm{2}}}\]

D. \[{{\rm{u}}_{\rm{k}}}{\rm{ = }}{{\rm{u}}_{\rm{1}}}{\rm{.}}{{\rm{q}}^{{\rm{k}} - {\rm{1}}}}\]

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