348 Demonstration of the Binomial Theorem for - [May, 
For r = 2 we have “J? + % = (a) 'I#*+2 pean = (‘? + It) 
p42 5 Pies ay (2—*) PTE yar cated Sk 112. TP otp 
5 = 'f ay) teas + I aml = Je + 
TI + °I, 
For r = 8 we have ‘J +? = eter san? (2) =(HP + °F. 
‘Ip 'la p —1 
3 
al —2 Pp 2qp ‘Iq sJp . 119 
29a), Fae yo 2 * 2]7 _ — sje Ea the) S92 ead Ya 2 
ee ee ee 
sJr *Y¢ 
yep ye) LETHE Lye PHF pp Dy + ‘IF. 
5 apd == = + He Ie + We elt + “Is, as it ought 
to be. 
In the same manner we may prove that 
Jot = sf + o]e ye te eye She ATs, 
But in order to generalize this induction, and to show that this 
must always be so, we shall suppose that, proceeding in this manner, 
we had convinced ourselves of its truth up to the values of 7, se 
that we had proved that 3 
sJe +q — sJp ae s—i]P fg de s— lyp Ya + s—3]p dL &e. 
And we shall prove that it will also be true for r = s + 13 for 
st ifota = spre potas = (substituting the value of ‘I+ ¢) 
(‘Ie + se Yo + s—2]p J see ‘Ty, &e. + YP t——, 
By multiplying every term by Bee, and writing the latter 
: —st+I1)+ —! = 9) + —_ 2 
successively eas+ eu, (p-s * : (q ) 
(p —s + 3) + (¢ — 3) - 
——_____——_-,, &c. we obtain 
s+l 
Je p-s s[p '¥¢a s—‘Ip Ia g—1 
. s+l s+l s+l 
s—lPp—s+l, s—2Ip, Iv p—s +2 
EE ae AE Re | | Bs aE Dee SATS «5 
s+l I? + s+ 1 
The first term is ‘ * ‘I? (2). We have also 
I, 319 wait = ‘Jp 
poh (Ses yy sO = Tb 
Fae, | ek so-F 
se tIsg—1 , .-P ps + 2 
hi, a= 0 Veco a. 
righ: atte SO Dirt Sideribe Ait ES IF 7 
s+1 “i 
And thus two terms produced by two successive terms will always 
give together the same terms in the product. Thus in the pro- 
ducts 
