6 PRACTICAL MATHEMATICS 



and since a, [3, y ... are derived only from A, B, C . . . they 

 must be independent of x and a. 



Thus (x + d) n can be written as a series of terms containing 

 descending powers of x and ascending powers of a, such that in 

 any one term the sum of the powers of x and a is always n. 



If we can find how the coefficients A, B, C . . . depend upon 

 n, then we can establish a general way of expanding (x + a) n . 



By multiplication 



A B C D E 



(x + a) 1 = x + a 

 (x + a) 2 = x z + 2ax + a 2 

 (x + a) 3 = x 3 + Sax 2 + 3a 2 x + a 3 

 (x + a)* = a? 4 + 4aa? 3 + 6a 2 x 2 + 4a 3 x + a 4 

 (x + a) 5 = x 5 + 5ax* + Wa 2 x 3 + I0a 3 x 2 + 5a*x + a 5 

 (x + a) 6 = x 6 + Gax 5 + 15a 2 ;r 4 + 20a 3 x 3 + 15a*x 2 + Qa 5 x + a 6 



The coefficients in the second vertical column give the different 

 values of A for corresponding values of n, and obviously A = n. 



As A depends upon the first power of n, let B depend upon the 

 second power of n, and the most general form that B can take 

 is b + en + dn 2 where b, c, and d are constants. 



Let B = b + en + dn 2 



whenn=2 B=l b + 2c + 4>d = 1 ... (1) 



whenw=3 B = 3 b + 3c + 9d = 3 ... (2) 



when n= 4> B=6 & + 4c + IQd =6 ... (3) 



Subtracting (1) from (2) c+ 5d = 2 . . . (4) 



Subtracting (2) from (3) c + 7d = 3 ... (5) 



Subtracting (4) from (5) 2d = I d = - 



from (4) c = 2 - 5d 



from (1) b = 1 -2c -4d b=0 



T,, n 2 n n(n 1) 



Thus B = - = - - 



If A= 



n - 1 



Then B = A 



X 



2 



That is, to obtain B from A, multiply A by the fraction whose 

 numerator is " one less " and whose denominator is " one more " 

 than that of A. 



If this rule for obtaining a coefficient from the preceding 

 coefficient is general, then to get C from B, C would be 



n-2 (n-l)(n-2) n(n-l}(n-2) 



B X r A X 2.3 r 1.2.3 



