New Algebraic Serves. 279 
1 lz <i _ 
+a) -+a(a+h)|-5 +a(a-+k)(a+-2h) 7 53th. 
+ 6] +-2ab* +Sab(a+h) AG cit 
+(b+k) | +3ab(b-+k) + 
+ b(b-+k)(b+-2k)| + 
Here the co-efficient of : is a+6; that of <> may be 
decomposed into a((a+k)+6)=a(a+6+k), and b((6-+4) 
+a)=b(+a+b+4), the sum of which is (a+b)(a+6+4). 
= 3 may likewise be decomposed 
#) (a+k)(a-+2k)+2b(a-+k)-+b(b6+4) 4 and | 
(b+k)(b-+2k)+2a(b-+k)+a(a+k) 7 
But the multiplier of a in the first part is evidently that 
The co-efficient of 
into 
Which would become the co-efficient of <i when a is 
changed into a+k, and the multiplier of 6, in the second, 
is the same co-efficient that would result by ces b in- 
to b+k. It follows, therefore, that, as the coefiicient of 
z : 
7.2» becomes (a+k)(a+b+h), the multiplier of a, in the 
* bg : 
first part of the co-efficient of 2? and that of bin the 
second, will become a(a+b+k)(a+b+2k)+b(a+b+h) 
(+642), that is (a+b)(a+s-+k)(a+b+2k). Hence, 
from the preceding, the law is evident for the four first 
terms. And to show that the law will hold for any number 
of terms of the product of the above series, we have onl: 
to prove that if the law holds as far as the co-efficient 
it will equally hold for the next co- 
Boson: Sat ge 
1.2...(p—t inclusively, 
= z* a 
* In the actual multiplication 4 X 7 =900- 1.92, 9" 
* 3 4 
3 2 
1 (a-+k) == 3ab(a+h) 79,39 &e- 98 above 
