Cho một cấp số nhân \(\left( {{u_n}} \right)\) có công bội \(q\) thỏa mãn \(\left\{ \begin{array}{l}{u_5} + {u_2} = 36\\{u_6} - {u_4} = 48\end{array} \right.\). Tính \({u_1} + 2024q\).

\(\left\{ \begin{array}{l}{u_5} + {u_2} = 36\\{u_6} - {u_4} = 48\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1}{q^4} + {u_1}q = 36\\{u_1}{q^5} - {u_1}{q^3} = 48\end{array} \right.\)\[ \Leftrightarrow \left\{ \begin{array}{l}{u_1}q\left( {{q^3} + 1} \right) = 36\\{u_1}{q^3}\left( {{q^2} - 1} \right) = 48\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}{u_1}q\left( {q + 1} \right)\left( {{q^2} - q + 1} \right) = 36\\{u_1}{q^3}\left( {q - 1} \right)\left( {q + 1} \right) = 48\end{array} \right.\]

$\iff \left\{\begin{array}{l}{u}_{1}q\left(q+1\right)\left(q^2-q+1\right)=36\\ \frac{36q^2\left(q-1\right)}{q^2-q+1}=48\end{array}\right.$ $\iff \left\{\begin{array}{l}{u}_{1}q\left(q+1\right)\left(q^2-q+1\right)=36\\ 3q^2\left(q-1\right)=4\left(q^2-q+1\right)\end{array}\right.$ $\iff \left\{\begin{array}{l}{u}_{1}q\left(q+1\right)\left(q^2-q+1\right)=36\\ 3q^3-7q^2+4q-4=0\end{array}\right.$ $\iff \left\{\begin{array}{l}{u}_1\cdot 2\left(2+1\right)\left(2^2-2+1\right)=36\\ q=2\end{array}\right.$ $\iff \left\{\begin{array}{l}{u}_1=2\\ q=2\end{array}\right.$

Vậy \({u_1} + 2024q = 2 + 2024 \cdot 2 = 4050\).

Trả lời: 4050.