Cho cấp số nhân (un) có \({S_2} = 4;\,{...

B

- Hướng dẫn giải

Ta có:

$\left\{ {\begin{array}{*{20}{c}} {{S_2} = \frac{{{u_1}\left( {1 - {q^2}} \right)}}{{1 - q}} = 4}\\ {{S_3} = \frac{{{u_1}\left( {1 - {q^3}} \right)}}{{1 - q}} = 13} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {{u_1}\left( {1 + q} \right) = 4\,\,\,\,\,\,\,\,\,\,\,\,}\\ {{u_1}\left( {1 + q + {q^2}} \right) = 13} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {\frac{{1 + q}}{{1 + q + {q^2}}} = \frac{4}{{13}}\,\,\,\left( 1 \right)}\\ {{u_1} = \frac{4}{{1 + q}}\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)} \end{array}} \right.$

Xét $\left( 1 \right):\,\,\,\frac{{1 + q}}{{1 + q + {q^2}}} = \frac{4}{{13}} \Leftrightarrow 4{q^2} - 9q - 9 = 0 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {q = 3 \Rightarrow {u_1} = 1}\\ {q = - \frac{3}{4} \Rightarrow {u_1} = 16} \end{array}} \right.$

Với $q = 3;\,{u_1} = 1 \Rightarrow {u_2} = {u_1}.q = 3 > 0$ (loại)

Với $q = - \frac{3}{4};\,{u_1} = 16 \Rightarrow {u_2} = {u_1}.q = - 12 < 0$ (Thỏa mãn).

Vậy ${S_5} = \frac{{{u_1}\left( {1 - {q^5}} \right)}}{{1 - q}} = \frac{{16\left( {1 - {{\left( { - \frac{3}{4}} \right)}^5}} \right)}}{{1 + \frac{3}{4}}} = \frac{{181}}{{16}}$.