2 2×2(v+1) G1=0 3(3+2V) 4(2v+4) C=0 于 (lCo C1=0 2n(v+m)Xv+-1)…(+) 1yco【v+) n=1,2;… 2n!r(v+n+D)∴ 0 2 1 3 2 4 5 - , 2 2 ( 1) - 0, 3( 3 2 ) - , 4 ( 2 4 ) 0 C C C C C C C n n n = ´ + = = + = + = 于是 0 2 2 2 1 0 2 (-1) ; 0 2 !( )( -1) ( 1) (-1) ( 1) 1,2, 2 ! ( 1) n n n n n n C C C nnn C n n n n n n n n = º + + + ××× + G + = = ××× G + + ；