Dn+1=( 1)
=( 1)=( 1)=
n(n+1)2
n+1≥i>j≥1
∏[(a i+1) (a j+1)]
n(n+1)2
n+1≥i>j≥1n(n+1)2
∏[ (i j)]
n+(n 1)+ +1
2
( 1)
n+1≥i>j≥1
∏(i j)
n+1≥i>j≥1
∏(i j).
an
(4)D2n=
bn
;
a1b1c1d1
cn
解
dnbn
(按第1行展开)
an
D2n=
cn
a1b1c1d1
dn
an 1
=an
bn 10
a1b1c1d1
cn 10dn 100dn