Cho cấp số nhân\[\frac{{\rm{1}}}{{\rm{2}}}{\rm{; }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{; }}\frac{{\rm{1}}}{{\rm{8}}}{\rm{ ;}}...{\rm{; }}\frac{{\rm{1}}}{{{\rm{4096}}}}\]. Hỏi số \[\frac{1}{{4096}}\]là số hạng thứ mấy trong cấp số nhân đã cho?

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

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. \[{\rm{128;}} - {\rm{64; 32;}} - {\rm{16;}} - {\rm{8; }}...\]

B. \[\sqrt {\rm{2}} {\rm{; 2; 4; 4}}\sqrt {\rm{2}} {\rm{; }}...\]

C. \[{\rm{5; 6; 7; 8; }}...\]

D. \[{\rm{15; 5; 1; }}\frac{{\rm{1}}}{{\rm{5}}}{\rm{; }}...\]

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

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

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

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

A. q = 2

B. q = – 2

C. \[{\rm{q = }} - \frac{3}{2}\]

D. \[{\rm{q = }}\frac{{\rm{3}}}{{\rm{2}}}\]

A. 56⁰

B. 102⁰

C. 252⁰

D. 168⁰