Tìm số hạn tổng quát của cấp số nhân

$\begin{array}{l}
\left\{ \begin{array}{l}
{u_2} - {u_4} + {u_5} = 10\left( 1 \right)\\
{u_3} - {u_5} + {u_6} = 20\left( 2 \right)
\end{array} \right.\\
\left( 2 \right) - 2.\left( 1 \right) \Rightarrow {u_3} - 2{u_2} + {u_5} - 2{u_4} + {u_6} - 2{u_5} = 0\\
 \Leftrightarrow {u_2}\left( {q - 2} \right) + {u_4}\left( {q - 2} \right) + {u_6}\left( {q - 2} \right) = 0\\
 \Leftrightarrow \left( {q - 2} \right)\left( {{u_2} + {u_4} + {u_6}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
q = 2\\
{u_2} + {u_4} + {u_6} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
q = 2\\
{u_1}\left( {q + {q^5} + {q^3}} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
q = 2\\
{u_1} = 0(L)\\
q + {q^5} + {q^3} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
q = 0(L)\\
{u_1} = 0(L)\\
{q^4} + q^2 + 1 = 0(L)\\
q = 2
\end{array} \right.\\
 \bullet q = 2,\left( 1 \right) \Leftrightarrow {u_1}.2 - {u_1}{.2^3} + {u_1}{.2^4} = 10\\
 \Leftrightarrow {u_1}\left( {16 - 8 + 2} \right) = 10\\
 \Leftrightarrow {u_1} = 1\\
 \Rightarrow \left( {{u_n}} \right):{u_n} = {2^{n - 1}}
\end{array}$