Cho hai số −3 và 23. Xen kẽ giữa hai số đã cho n số hạng để tất cả các số đó tạo thành cấp số cộng có công sai d = 2. Tìm 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) \(\left\{ \begin{array}{l}{u_4} - {u_2} = 36\\{u_5} - {u_3} = 72\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1}{q^3} - {u_1}q = 36\\{u_1}{q^4} - {u_1}{q^2} = 72\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1}q\left( {{q^2} - 1} \right) = 36\\{u_1}{q^2}\left( {{q^2} - 1} \right) = 72\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 6\\q = 2\end{array} \right.\).

b) \({u_6} = {u_1}{q^5} = 6 \cdot {2^5} = 192\).

c) \({S_5} = \frac{{{u_1}\left( {1 - {q^5}} \right)}}{{1 - q}} = \frac{{6\left( {1 - {2^5}} \right)}}{{1 - 2}} = 186\).

d) Ta có \({u_n} = {u_1}{q^{n - 1}} = 6 \cdot {2^{n - 1}}\).

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