Bn(=)=Bn1(=)+kmBn1(=)(1) Bm(=)=kmBm-(= Bn1(-)=Bn(=) Ek B1(=)(3) (3)代入(1)得(4) Bn()=12[Bn()-knBn()(4) (4)代入(3)得: B -zkm Bn()2Bm ()(3)代入(1)得(4) 1 2 1 4 1 m m m m m B z B z k B z k 1 1 1 1 1 1 1 2 m m m m m m m m B z B z k z B z B z k B z z B z Bm1 z zBm z zkm Bm1 z 3 (4)代入(3) 得: 1 2 1 1 m m m m m B z zk B z zB z k