Mr. Sewell's Demonstration, &c. 
3 8 3 
numerator and denominator of the fraction, respectively ter- 
minate in m and n terms. Suppose then x =y ; then will 
v — z; and our equation will become , or ” — = A 
-f 2 Bx -f- 3 Cx z + 4 Djc 3 & c. 
But v n = i -j- therefore by multiplying we have ~~~ A 
-f A -f- 2 B.r -f- eB S^x z -f- 3C -f- 4D.2: 3 +> &c. Or if = 
m 
r+^F= + 2^2 a + 2,,B - f3,lC z* + J22±*^.x>+,&c. 
Compare this with the assumed series, to which it is similar 
and equal, and it will be 
nA __ 
m 
nA-\- 2«B . 
m 
2?zB-f 3«C 
rn 
&C. =, &C. 
B, 
A=”; B 
n 7 
n A 
; C 
1.2.3 .n 
; &c. 
Therefore 1 -j- x\ n = 1 -j- — .r -f 
mxm — nxm — 2« 
I.Z.3.H 3 
&c. the law is manifest, and agrees with the common 
form derived from other principles. 
Sch. In the above investigation, it is obvious that unless m 
be a positive whole number, the numerator abovementioned 
does not terminate : it still remains, therefore, to shew how to 
derive the series when m is a negative whole number. In this 
case, the expression ( if — z m ) assumes this form, £ — or 
its equal which divided by v n — z n , as before, gives 
, &c. = 
—V—ZX. :v m ~ I -\-v m — z Z+V m ~ 3 z* 
" z m V”— z n 
MDCCXCVl. 
v — X : v n • 
3 D 
— a z+V n — 3 Z z + 
