Cho \(\left( {{u_n}} \right)\) là cấp số cộng có \({S_n} = {n^2} + 4n\) với \(n \in {\mathbb{N}^*}\). Số hạng đầu \({u_1}\) và công sai d của cấp số cộng đó là:

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

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

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

Trả lời: 4050.

Ta có \(\left\{ \begin{array}{l}{u_3} =  - 3\\{u_8} =  - 23\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} + 2d =  - 3\\{u_1} + 7d =  - 23\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 5\\d =  - 4\end{array} \right.\).

Khi đó \({S_{16}} = \frac{{16}}{2}\left( {2{u_1} + 15d} \right) = 8\left[ {2 \cdot 5 - 15 \cdot \left( { - 4} \right)} \right] = 560\).

Trả lời: 560.