Demonstration of the Binomial Theorem. 133 
=o+ ms ced lea) Fe a 4. m x (m—1) x (m—2) ™-3 
ax* 
oy anes a 1x2x3 at 
&c. ad infin. i — m=1, and I have put m instead si 
in order to avoid confusion in the process. 
Now multiply both sides of the equation (a+ +2) =¢ a+ m nar £ 
mx (m—1) "%, mx (m—1)x(m—2)_ ™P j 
——— ax 7; oo ax*+ &c. ad inf. 
by a+z and I have (a+2)x(a+2)= (a+r) = atm a fot 
mer meer pskéere)) (m— 2) 
1x 
4+}xa% mee ? 
ad infinitum. 
xa te? +e. ad inf. 
mxX(m—] 
sa tt eicrc ae Kaxt4 Be, 
+1 m 
Hence by adding hee = «a pm la r+ (m+ 1) 1) x 
os ¥ 
m—t1 
a «?+4+(m+1)x(m)xm—1 =2 
(m) +( ane 3 a «*+48&c, ad infin. which 
is of the same form as the expression for (a ta)” only m+1 
takes the place of m, that is, m is every where increased by 1. 
In like manner, if both sides of this equation are multiplied 
by a+a, m+1 will be increased by 1, or will be changed 
into m+2, and so on to m+3, m+ 4, m+n, n, denoting any 
positive integer which gives (a+: es is ins 7 
a Ss 
m+-n—2 
4+(m+n)x(m4+n—1) a 274+(m+n)x (m+ _ 
1x2 x 
mt+tn—3 >? 
nm—1)x(m4+n—2) a 2° + &e. ad infin. or (a+z) 2a + 
2xs a 
p-l p22 ges: 
pazr+px(p—l1) a x? +p X(p—1) x (p—2) aE &e. 
= Exe Ix2xS 
