Đáp án:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
{u_1} + {u_2} + {u_3} + {u_4} = 30\\
{u_5} + {u_6} + {u_7} + {u_8} = 480
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}\left( {1 + q + {q^2} + {q^3}} \right) = 30\\
{u_1}.{q^4}\left( {1 + q + {q^2} + {q^3}} \right) = 480
\end{array} \right.\\
\Rightarrow {q^4} = 480:30 = 16\\
\Rightarrow q = 2/q = - 2\\
\Rightarrow {u_1} = 2/{u_2} = - 6\\
\Rightarrow CSN:\left[ \begin{array}{l}
{u_n} = {2^n}\\
{u_n} = \left( { - 6} \right).{\left( { - 2} \right)^{n - 1}} = - 3.{\left( { - 2} \right)^n}
\end{array} \right.\\
b)\left\{ \begin{array}{l}
{u_5} - {u_1} = 15\\
{u_4} - {u_2} = 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}\left( {{q^4} - 1} \right) = 15\\
{u_1}\left( {{q^3} - q} \right) = 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{u_1}\left( {{q^2} - 1} \right)\left( {{q^2} + 1} \right) = 15\\
{u_1}.q.\left( {{q^2} - 1} \right) = 6
\end{array} \right.\\
\Rightarrow \frac{{{q^2} + 1}}{q} = \frac{{15}}{6}\\
\Rightarrow 6{q^2} - 15q + 6 = 0\\
\Rightarrow q = 2/q = \frac{1}{2}\\
\Rightarrow {u_1} = 1/{u_1} = - 16\\
\Rightarrow CSN:\left[ \begin{array}{l}
{u_n} = {2^{n - 1}}\\
{u_n} = \left( { - 16} \right).\frac{1}{{{2^{n - 1}}}} = \left( { - 1} \right).\frac{1}{{{2^{n - 5}}}}
\end{array} \right.
\end{array}$
