Cho cấp số nhân (u_n) có {u1 + u2 = 3; u3 + u4 = 12, biết rằng q > 0. Tính S4.

Chọn B
\(\left\{ \begin{array}{l}{u_1} + {u_2} = 3\\{u_3} + {u_4} = 12\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1} + {u_1}q = 3\\{u_1}{q^2} + {u_1}{q^3} = 12\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1}\left( {1 + q} \right) = 3\\{u_1}{q^2}.\left( {1 + q} \right) = 12\end{array} \right.\)
\( \Rightarrow \frac{{{u_1}{q^2}.\left( {1 + q} \right)}}{{{u_1}\left( {1 + q} \right)}} = \frac{{12}}{3} = 4 \Leftrightarrow {q^2} = 4 \Leftrightarrow q = 2\) (do \(q > 0\))
\({u_1} = \frac{3}{{1 + q}} = \frac{3}{{1 + 2}} = 1\).
Suy ra \({S_4} = {u_1}.\frac{{1 - {q^4}}}{{1 - q}} = 1.\frac{{1 - {2^4}}}{{1 - 2}} = 15\).