回復 1# justhgink 的帖子
\( a^2_{n+1} +2 \cdot a_{n+1} = 4 \cdot S_{n+1} \)
\( a^2_{n} +2 \cdot a_{n} = 4 \cdot S_{n} \)
相減得
\( a^2_{n+1} +2 \cdot a_{n+1} -( a^2_{n} +2 \cdot a_{n}) = 4( S_{n+1} - S_n ) = 4 a_{n+1} \)
得
\( (a_{n+1} + a_n )(a_{n+1}-a_n -2)=0 \)
若 \( a_n >0 \)
則 \(a_{n+1}- a_n =2 \) 等差
[ 本帖最後由 cplee8tcfsh 於 2012-5-5 09:21 PM 編輯 ]