124 4. Particular Determinants
Identities 1.
Vnr=
n
∑
j=1
K
rj
n,^1 ≤r≤n. (4.10.4)
Vnr=
(−1)
n+r
(h+r+n−1)!
(h+r−1)!(r−1)!(n−r)!
, 1 ≤r≤n. (4.10.5)
Vn 1 =
(−1)
n+1
(h+n)!
h!(n−1)!
. (4.10.6)
Vnn=
(h+2n−1)!
(h+n−1)!(n−1)!
. (4.10.7)
K
rs
n =
VnrVns
h+r+s− 1
, 1 ≤r, s≤n. (4.10.8)
K
r 1
n =
VnrVn 1
h+r
. (4.10.9)
K
nn
n
=
Kn− 1
Kn
=
V
2
nn
h+2n− 1
. (4.10.10)
K
rs
n
=
(h+r)(h+s)K
r 1
nK
s 1
n
(h+r+s−1)V
2
n 1
. (4.10.11)
Kn=
(n−1)!
2
(h+n−1)!
2
(h+2n−2)!(h+2n−1)!
Kn− 1. (4.10.12)
Kn=
[1!2!3!···(n−1)!]
2
h!(h+ 1)!···(h+n−1)!
(h+n)!(h+n+ 1)!···(h+2n−1)!
. (4.10.13)
(n−r)Vnr+(h+n+r−1)Vn− 1 ,r=0. (4.10.14)
Kn
n
∏
r=1
Vnr=(−1)
n(n−1)/ 2
. (4.10.15)
Proof. Equation (4.10.4) is a simple expansion ofVnrby elements from
rowr. The following proof of (4.10.5) is a development of one due to Lane.
Perform the row operations
R
′
i
=Ri−Rr, 1 ≤i≤n, i=r,
onKn, that is, subtract rowrfrom each of the other rows. The result is
Kn=|k
′
ij
|n,
where
k
′
rj=krj,
k
′
ij=kij−krj
=
(
r−i
h+r+j− 1
)
kij, 1 ≤i, j≤n, i=r.
After removing the factor (r−i) from each rowi,i=r, and the factor
(h+r+j−1)
− 1
from each columnjand then cancelingKnthe result can