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(1) 有
\[ a_{n+1}+(-1)^n a_n=An+B \]
及
\[ a_{n+2}+(-1)^{n+1} a_{n+1}=A(n+1)+B \]
两式消去$a_{n+1}$得
\[ a_{n+2}+a_n=A(n+1)-(-1)^{n+1}An+B[1-(-1)^{n+1}] \]
于是
\[ a_{n+4}+a_{n+2}=A(n+3)-(-1)^{n+3}A(n+2)+B[1-(-1)^{n+3}] \]
两式再消去$a_{n+2}$得
\[ a_{n+4}-a_n = 2A-(-1)^{n+1}\cdot 2A \]
所以
\[ a_{4n+5}-a_{4n+1}=4A, a_{4n+6}-a_{4n+2}=0, a_{4n+6}-a_{4n+3}=4A, a_{4n+7}-a_{4n+4}=0 \]
下略. |
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