Cho cấp số nhân \(\left( {{u_n}} \right)\) có \(\left\{ \begin{array}{l}{u_4} + {u_6} = - 540\\{u_3} + {u_5} = 180\end{array} \right.\) . Tính S21.
Giải chi tiết:
\(\begin{array}{l}\left\{ \begin{array}{l}{u_4} + {u_6} = - 540\\{u_3} + {u_5} = 180\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1}{q^3} + {u_1}{q^5} = - 540\\{u_1}{q^2} + {u_1}{q^4} = 180\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1}{q^3}\left( {1 + {q^2}} \right) = - 540\\{u_1}{q^2}\left( {1 + {q^2}} \right) = 180\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}q = - 3\\{u_1} = 2\end{array} \right.\\ \Rightarrow {S_{21}} = \frac{{{u_1}\left( {1 - {q^{21}}} \right)}}{{1 - q}} = \frac{{2\left( {1 - {{\left( { - 3} \right)}^{21}}} \right)}}{{1 - \left( { - 3} \right)}} = \frac{1}{2}\left( {1 + {3^{21}}} \right)\end{array}\)
Chọn A.