Demonstration of the Binomial Theorem. 135 
ite hate pad kip 4 
(@+e)aa +(+P) ae <= x (+ pal) a2? + 
1 1x2 
+pq 
&c. ad infin. from es) “tp + me &e. ad. 
infin. we must evidently divide + 33 wherever it occurs 
ie 
by g. Hence it appears, that if I have the equation (a+2) 
+r +r—-—1 +r—2 
—@+(+r) a «+(4+r)x(+r—-1)4 a*+ &e. ad in- 
i 1x2 
fmitum, and wish to obtain the s root of it, I must nape 
+r wherever it occurs, by s, and that whether ctr be ex 
ly divisible by s, or not. For as the same rule is used in the 
Se ea? ret 
extraction of the s root of G+r)— a+ + (+r) ax + 
gee 
&e. ad infin. as in the extraction of the qg root of (a+ x) 
+pq + +79g-1 | — 
sit ale +pq)a_ x + &c. ad infin. and as the q root is 
had by dividing +pg wherever it occurs, by q : therefore 
a) ace a 
by — - process, the s root of (a+z) = @ + 
(ena we we &e. bs see ee ehyee which 
ia 3 Surg 
gives (a4 = s+(+ — fe eer +9) x (+ 1) 
s 
gees “1x2 
ee, are &e. sdinf whichis yet ofthe same form as when te 
exponent is integral. Hence, universally, I have (a+z) Han % 
a sagscla _1) az? #2 &c. ad infin. sa ans u denotes any 
ssamibey Paliaer. whether integral or fractional, posers or 
