THE BINOMIAL THEOREM.] 



MATHEMATICS. ALGEBRA. 



615 



begin from a or from x, we get the same coefficients. 

 Thus 



a? + 8x'a + 28x B a + 56x 5 a 3 + 70x*a* -\ 



+ 28x 2 a 6 + 8xa 7 + <# 

 a* + 8a 7 x + 28a 6 x 2 + SCaV + 70a*x* + 



+ 28a 2 x + 8ax 7 -f- x s . 



It can easily be seen that this must be the case, for 

 the former of these two is (x + a)", and the latter 

 (o + x)', which are clearly the same thing. This con- 

 sideration greatly facilitates our expansion of a binomial 

 Thus 



10"9 10"9'8 - i 



(a + x)' = a' + 10 a'x + ^-aV + _ * 



10-9-8-7 



10-9-8 -7 -6 



- 1200,'x 3 + 210ax 4 

 + 210ax 6 '+ WOaV + 45aV + lOax 9 + z 10 . 

 In like manner 

 (a + x) 9 = a 9 + 9ax + 36a 7 :r 2 + 84ax -f 121ax* 



+ 12]a*x B + 84a 3 x -f- 36a 2 x 7 + 9ax s +x 9 . 

 It will also be observed that, in the case of (a + x) 1 

 there is a middle term, 252a 5 x 5 ; but in the case of 

 (o + x) 9 there is no middle term. In fact, it is plain 

 that the expansion of (a + 6)" contains n -f 1 terms, for 

 the expansion contains one term in which 6 does not 

 appear, and also terms containing 6, 6 2 , 6 s .... up to 

 6". Hence, if n be even, n + 1 is odd, and the expansion 

 has a middle term ; if n be odd, n + 1 is even, and the 



usion is without a middle term. 



10. To prove the Binomial Theorem for any value of n. 

 Let us use the notation 



1) 



for all values of n. 

 and 



<fcc. 



Now, when m and n are positive integers, we obtain 

 from equation (1 ) 



m.(m 1) 



n)(m + n 1) 

 ~ ~ x 



n.(n 1) 

 -!- 



n 1) 

 1 



. . .) X 



= (1+ 



Hence, when m and n are any positive integers what- 

 ever, 



/(m)/(n)=y(7n + n) 



In accordance with the principle of the permanence of 

 equivalent forms, we assume this to hold good, whatever 

 m and n are, and then interpret the meaning we must 

 assign to /(m) consistently with this assumption. 



It in plain that 



f(m) /(n) f(f) - flm + n) flp) = f(m + n + p), 

 nd so on for any number of terms. 



Hence, 



/(P) X /(?) X/?) toq factors 

 fractions) /. 





+ to 2 



Now p is a positive integer, 

 .-. 1 +px +P^S) 3 



.../(P) = (l + x)f 



Let? =n 

 2 



when i is a fraction. 

 Again, if n be negative we have 



A") Am) /( m) = /(n + 



/W A- )- 



m) =/(n) 



Now, m is positive . . /(m) = (1 + x) 



/(m) 



Let n = m 



( m). 



when n is negative. 



And therefore, for all values of n positive or negative, 

 integral or fractional, 



(1 + x). - 1 + nx + ^ 



x*+.... (in.) 



Letx-- 



Now, o- 



*=* 5 + 



which is the Binomial Theorem, and we have proved it 

 to hold good for all values of positive or negative, 

 fractional or integral. 



N.B. The Binomial Theornm was discovered by 

 Newton ; it is very important, and is constantly used in 

 almost every part of mathematics ; it is therefore very 

 necessary that the student should be quite familiar with it. 

 It will be well if he will carefully examine the following 

 results. 



(1.) To show that 



n.(n-I) n (n-l)(n-2) 



(1 x)'-l n.x+ ~ 7 x - 



We obtain this from equation (IV.) by writing o = 1, 

 and b = x. 



(2). To show that 



n-(n-l) n-frv-1) (n-2) 

 2 = 1 1- n + t . a T2 -3~~ 



write in the series (IV.) o = 1 and 6 = 1, and we obtain 

 this result. 



(3). To show that 



n(n 1) (n. 2) n(n 1) . 



n -4- ?~s Q r <fc' = 1 T i~.o r &- 



1-2 o I 4 



write in the series (IV.) a = 1 and 6 = 1. Then 

 n(n 1) n.(n 1] 



1-2-3 



-^_,fco. 



Whence the result. 



