a) \(\left\{ \begin{array}{l}{u_3} = 15\\{u_5} = 135\\{u_6} < 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_3} = 15\\{u_3}{q^2} = 135\\{u_6} < 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_3} = 15\\{q^2} = 9\\{u_6} < 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1}{q^2} = 15\\{q^2} = 9\\{u_6} < 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} = \frac{5}{3}\\q = - 3\end{array} \right.\).

Suy ra \({u_n} = \frac{5}{3} \cdot {\left( { - 3} \right)^{n - 1}}\).

b) \(\left\{ \begin{array}{l}{u_1} + {u_5} = 164\\{u_2} + {u_3} + {u_4} = 78\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} + {u_1}{q^4} = 164\\{u_1}q + {u_1}{q^2} + {u_1}{q^3} = 78\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1}\left( {1 + {q^4}} \right) = 164\\{u_1}q\left( {1 + q + {q^2}} \right) = 78\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} = \frac{{164}}{{1 + {q^4}}}\\\frac{{164}}{{1 + {q^4}}}q\left( {1 + q + {q^2}} \right) = 78\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} = \frac{{164}}{{1 + {q^4}}}\\82q\left( {1 + q + {q^2}} \right) = 39\left( {1 + {q^4}} \right)\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} = \frac{{164}}{{1 + {q^4}}}\\82q + 82{q^2} + 82{q^3} = 39 + 39{q^4}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} = \frac{{164}}{{1 + {q^4}}}\\\left( {q - 3} \right)\left( {q - \frac{1}{3}} \right)\left( {39{q^2} + 48q + 39} \right) = 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} = \frac{{164}}{{1 + {q^4}}}\\\left[ \begin{array}{l}q = 3\\q = \frac{1}{3}\end{array} \right.\end{array} \right.\).

Với \(q = 3\)\( \Rightarrow {u_1} = 2\). Khi đó \({u_n} = 2 \cdot {3^{n - 1}}\).

Với \(q = \frac{1}{3} \Rightarrow {u_1} = 162\). Khi đó \({u_n} = 162 \cdot {\left( {\frac{1}{3}} \right)^{n - 1}}\).

Cho các dãy số sau, dãy số nào là dãy số vô hạn?