484 Mr WARBURTON, ON SELF-REPEATING SERIES. 



Hence if from 2 n+v we subtract the equals of the equation (26), and divide the left side 

 of the resulting equation by 2, we shall obtain the equation, 



n + l-v ra+2-i) 



^ [the coefficient of A0 2n + 1 -the middle term of that coefficient] = the coefficient of A0 2n+1 ; (27) 



M + i-o l~n + l~| 



and the middle term of the coefficient of A0 2n+1 is the term — — r= — , or the (v + l) th term 



IV 1 1 ' V ■ / 



of the figurate series of the « th order. 



ra + 2 



But in Diagram IV., when v = 0, the coefficient of A0 2n+1 = 0; and when v = 0, 

 b|+1 



(n + 1) n + 1 



i — jTj — =1; and therefore the coefficient of A0 2n + 1 = l. Consequently, 



n 



the coefficient of A0' 2 " + 1 = (n + l) + l x 2. 



2| + 1 

 nfU 



n - 1 \n a. 1 1 



A0 2„ + i = k_!_J + ( n + i) x 2 + ! x 2 2 



3|+1 2| + 1 



n-2 



A0 2 » +1 = ***£* + ^4lT ^ 2 + ( w + X 2 2 + 1 x 2 3 . . (28) 



n| + l n-l| + l 



A0 1 



1 „ , fra + ll \n + ll 



^ 2«+i = 1 __J_ + l — jL x2 + + (w + l)2"" 1 + 1 x2" = 2 8 ". 



And thus, when 2ra + 1 = 9, 

 the coefficient of A'O 9 = 70 x 1 + 35 x 2 + 15 x 2 2 + 5 x 2 3 + l x 2* = 2 8 . 



A 2 9 = 35 x 1 + 15 x 2 + 5 x 2 s + 1 x 2 :) 



A s 9 = 15 +5x2+1x2' 



A 4 9 = 5x1 +1x2 



A 5 9 = lxl. 



And hence the following ready mode of determining these coefficients : 



the coefficient of A'O 9 = 256. 



._ 256 - 70 

 A ? 9 = - = 93. 



A 4 9 = 



