Trong các dãy số \({u_n}\) cho bởi số hạng tổng quát \({u_n}\) sau, dãy số nào bị chặn?

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

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

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

Trả lời: 4050.

a) Ta có \({u_2} = {u_1} + 5 = 2 + 5 = 7\).

b) Có \(d = {u_{n + 1}} - {u_n} = 5\).

c) Ta có \({u_n} = {u_1} + \left( {n - 1} \right)d = 2 + \left( {n - 1} \right) \cdot 5 = 5n - 3\).

d) Ta có \({S_{10}} = 10{u_1} + \frac{{10 \cdot 9 \cdot 5}}{2} = 20 + 225 = 245\).

\({S_{100}} = 100{u_1} + \frac{{100 \cdot 99 \cdot 5}}{2} = 200 + 24750 = 24950\).

Vậy \(S = {u_{11}} + {u_{12}} + ... + {u_{100}} = {S_{100}} - {S_{10}} = 24705\).

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