Cho cấp số cộng \(\left( {{u_n}} \right)\) biết \(\left\{ \begin{array}{l}{u_1} = 2\\{u_{n + 1}} = {u_n} + 5\end{array} \right.\) với \(n \ge 1,n \in \mathbb{N}\).

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

\[ \Leftrightarrow \left\{ \begin{array}{l}{u_1}q\left( {q + 1} \right)\left( {{q^2} - q + 1} \right) = 36\\\frac{{36{q^2}\left( {q - 1} \right)}}{{{q^2} - q + 1}} = 48\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}{u_1}q\left( {q + 1} \right)\left( {{q^2} - q + 1} \right) = 36\\3{q^2}\left( {q - 1} \right) = 4\left( {{q^2} - q + 1} \right)\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}{u_1}q\left( {q + 1} \right)\left( {{q^2} - q + 1} \right) = 36\\3{q^3} - 7{q^2} + 4q - 4 = 0\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}{u_1} \cdot 2\left( {2 + 1} \right)\left( {{2^2} - 2 + 1} \right) = 36\\q = 2\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 2\\q = 2\end{array} \right.\].

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

Trả lời: 4050.

a) Ta có \({u_4} = {u_2}{q^2}\)\( \Leftrightarrow 9 = 4{q^2}\)\( \Leftrightarrow q =  - \frac{3}{2}\) vì \(q < 0\).

b) \({u_1} = \frac{{{u_2}}}{q} = \frac{4}{{\frac{{ - 3}}{2}}} =  - \frac{8}{3}\).

c) \({u_5} = {u_4}q = 9 \cdot \frac{{ - 3}}{2} =  - \frac{{27}}{2}\).

d) \({u_8} = {u_1}{q^7} = \frac{{ - 8}}{3} \cdot {\left( {\frac{{ - 3}}{2}} \right)^7} = \frac{{729}}{{16}} \ne \frac{{ - 2187}}{{32}}\).

Đáp án: a) Đúng;    b) Đúng;   c) Sai;   d) Sai.