KiRSTiNE Smith 
35 
] 
1 
+ a 
+ a 
1 
2?+ 1 
1 
2^+3 
+ a 
+ a 
+ a 
1 
1 
2q + 2p - 1 
+ a 
+ a 
- + a 
\ 2q + 2p-3 2q + 2p-l 2q + 4p - 5 
By subtracting from the elements of each row the elements of the proceeding 
and leaving the first row as it is, it is transformed to 
] 
o 1' + ^ 
2q — 1 
2 
2?+ 1 
2 
+ a 
2q + 2p-3 
2 
+ a 
(2<7-l)(2g+l) 
;2g-M)(2g + 3) 
2 
(2^ + 2^ - 3) (2^+225-1) 
2 
Q 
,3 = 
{2q + 2p - 5) (2^ + 2p - 3) {2q + 2p - 2>) {2q + 2p - I) {2q + ip - 7) (2? + ip-b) 
which when the columns undergo the same process takes the form 
+ a 
2q - 1 
2 
(2^-1) (2^ + 1) 
2 . 4 
■2q+l){2q + ^) 
2.4 
\2q + 2p-b) {2q + 2p-3) 
2.4 
{2q - 1) (2g + 1) (2^ - 1) {2q + 1) (2g + 3) {2q + 1) (2^ + 3) {2q + 5y {2q+2p-5)...{2q+2p-l) 
2.4 
2.4 
2.4 
(2^+1) {2q + 3) {2q + 1) (2^ + 3) {2q + 5) {2q + 3) (2g + 5) {2q + ly {2q+2p-3)...{2q+2p+l) 
2 
2.4 
2 . 4 
2.4 
{2q+2p-5){2q+2p-d) i2q+2p-5)...{2q+2p-l) {2q+2p-^)...{2q+2p+iy {2q+4:p-9)...{2q-ip-5) 
Let us introduce the notation 
1 1 1 
3 
(2g-l) (2^+1) (2^ + 3) 
1 
(2^+1) (2^ + 3) (2(7 + 5) 
1 
(2^+1) (2^ + 3) (2^ + 5) 
1 
(2g + 3) {2q + 5y(2gTT) 
(2g+ 22^-3) ... (2g + 2^+ 1) 
1 
{2q + 2p-l) ...{2q+2p + 3) 
1 
{2q + 2p-3)... {2q + 2p + 1) {2q + 2p - I) ... {2q + 2p + 3) ^ i2q + Ap - 5) ... (2q + ip - 1) 
Then, since for a = 0 j,S equals the determinant j,A, we have 
,8=^A + a.23(f-i'.„_,Z) 
■(42), 
and the problem is reduced to the evaluation of j,Z). 
3—2 
