Tổng của 9 số hạng đầu tiên bằng
Đáp án A
Ta có: \(\left\{ {\begin{array}{*{20}{l}}{{u_4} - {u_2} = 54}\\{{u_5} - {u_3} = 108}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{u_1}{q^3} - {u_1}q = 54}\\{{u_1}{q^4} - {u_1}{q^2} = 108}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{u_1}q\left( {{q^2} - 1} \right) = 54}\\{{u_1}{q^2}\left( {{q^2} - 1} \right) = 108}\end{array}} \right.} \right.} \right.\)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{u_1} = \frac{{54}}{{q\left( {{q^2} - 1} \right)}}}\\{\frac{1}{q} = \frac{{54}}{{108}}}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{u_1} = \frac{{54}}{{2\left( {{2^2} - 1} \right)}}}\\{q = 2}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{u_1} = 9}\\{q = 2}\end{array}} \right.} \right.} \right.\).
Ta có: \({S_9} = \frac{{{u_1}.\left( {1 - {q^9}} \right)}}{{1 - q}} = \frac{{9.\left( {1 - {2^9}} \right)}}{{1 - 2}} = 4599\).