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IP属地:广东来自iPhone客户端1楼2014-04-21 19:31回复
    记:
    \[na_1+(n-1)a_{2}+(n-3)a_3+...+2a_{n-1}+a_{n}=T_{n}\]
    \[n^2a_1+(n-1)^2a_2+(n-3)^2a_3+...4a_{n-1}+a_{n}=Q_{n}\]
    \[n^3a_1+(n-1)^3a_2+(n-3)^3a_3+...8a_{n-1}+a_{n}=R_{n}\]
    \[S_{n}=\sum a_{n}\]
    则有:
    \[T_{n}+S_{n}+a_{n+1}=T_{n+1}\]
    \[Q_{n}+2T_{n}+S_{n}+a_{n+1}=Q_{n+1}\]
    \[R_{n}+3Q_{n}+3T_{n}+S_{n}+a_{n+1}=R_{n+1}\]


    IP属地:广东2楼2014-04-21 19:44
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