104 THE ROYAL SOCIETY OF CANADA 
1 m 
y+2r—1...v+2r— ete En TE ——— if = 
m 
a: SE aes vt+n—m—2 
ge toe 1...m+1 
a Er EU Er =m (i) 
1 À m— 1 1 
y+2r—1...v+2r—m—-1 & = 1...v+2r—m—2 
m—1 
a ) yy —2.. y Er mr 8 Fe 
= (—1)"-! DA DS SEE ty “a a 
p+2r—1...v+2r—2m m (ii) 
and (i) +m (ii) gives the required formula (12). 
§ 6. We shall next prove the identity 
ACDC st 1 y+1 
25+1\2+1. 2v+3.. Pp Dea 1.2y+1.y+1.94+2...y+54+1 
a Ny Doté aol ae nt eae ee Pat 
Co . 2r+8 .v+2.v+8...v+s+2 
Ss v+5 
ey pen em (13) 
We begin by converting 
25+1 
Bt eV Ur ell LE u+r+2...u+v+2s+1 
into partial fractions. It thus takes the form 
Eas 1 
Tae ESS EEE rae = ee 
(—1) 1 1 1 
DD OA es 0 ED pe se eee Les eee 
a (pp 1 1 = Saree 
2.4.2.4...2s—4 v+3.7+4...y+5+3 Lu—v—5 u+r+5 
ere: 
Multiply both sides by 2° . s!, put u=v, divide by 2(—1)s+1 and 
1 pide 
PRÉDIT by Tri and the result takes the form (13). 
It is worthy of notice that (12) can be readily derived from (13) 
For 
DRE LAS DES v+ 1 
p2—p+ 12. p2—y +32... w—v 25+ 1? we—(v+1)? v+1...v+5+1 
replace 
RDS VO eee See 
=G see —(v+3)? +2 .v+3 . ep ee ae NAN DRE 
v+ 5 
ao Le ee ee ee 
