Partial Fraction Problem. 



563 



By a similar method it can be shown that (3) holds also 

 for odd values of: k. 



It follows from (2) that A is formed in the same manner 

 as Q ln _i (if it existed). 



Therefore 



Pi- 1 ] 



71-1-2/3 



A = l {-l)W ~i (-i)v("t*) 



/8=0 a=0 \ P / 



m^-oB-x-i 



and 



p? 2 : 



h (ii) 



71-2-2)3 



B = m JI -/» 2 (-l)W 2 (-l)V 



<* + /3 



}»'t/i-2^-a-^ 



Again, let 2 t^jc^ -4- * denote the quotient and Aj.i'4-Bj 



ic=0 



the remainder (which will, at most, be linear) obtained by 



dividing X Qj K ^- 2 - K by j; 2 + aar4 6, then 



K-=0 



»-2 



2 Q u a* 



n-4 



2 Q 2K ^- 4 - K 



-+ 



A 1 .r + B 1 

 (.r + a^ + ^/f- 1 (** + aj? + 6)*-* ' (^-fa^ + 6)?- 1 ' l 1 -; 



where Q 2k is the same function of Q llc that Q lK is of m K . 



Clearing of fractions and equating coefficients of like 

 powers of as, we have 



Q2 K = Qi*— aQ 2 <c-i~6Q 2lc _ 2 (/c = 0, 1, 2, ..., n~2)^ 

 Ai = Qi»_ 3 — aQ 2n _ 4 — &Q 2K _ 5 , I (ify 



^l = Qln~2—0Q 2 n-4,- J 



We then obtain 



Q 2 o=w ; Q 21 =m 1 — 2am ; Q 22 =m 2 —2am l + 3a?m -2bm ; 

 Q 23 = m d — 2a m 2 4- 3a 2 m! — 4a 3 m — 2&(ro! — 3am ) ; 



Q24 = »U - 2am 3 -{- 3a 2 m 2 - ±a*m x + 5a 4 w? - 2b{m - 3an h + 4.a 2 m ) + 3b 2 m . 

 We now assume 



Q«J(-i,».C^)f ( - 1) ,,c+et'),,..„_.. (U) 



We shall now prove that this form is true for all values 



of fC. 



2 P 2 



