447 
whence by repetition, and noting (3) 
k! 
—l,27,k,0) = - 
Bt FO) a 1489), GEE) 
and 
vay "| a (—a2)" k! <a 
ASE id ioe cna ) aE (x+k) ie ee ae 
therefore, since by (10), f(—1,2,k,k) = —x f(—1,2,k,k—1), 
(12) f(—1,2,k,n) = (eevag n=0,1,2,3..k, but notn>k. 
att ay 7A Cre) eaereee (x+k) eH 4 
Example: 
4 , 
a(e-+1) (+2) (e+3) (+4) 2 (—1)' | ae = wu when n = 0 
i=0 a: 
= yh n= 1 
= 247? n=2 
= —24z3 n= 3 
= D4r ee 
but = 24024 + 840x? + 12002? + 5162, n = 5 
To find the value of f(—1l,2,k,n) for n > k, set m = 1 in (9) and multiply 
through by 
| 
(x+1)(a+2).... (wt+k)/S(k,k) = > BU k)x* /S(k,k) 
i=0 
and set 
k 
g(—1,z,k,n) for f(—1,2,k,n) > BG,k)x* */S(k,k): 
— 
k 
g(—1,2,k,n+1) = A(k,n) > B(i,k)x* '— xg(—1,2,k,n) 
= 
IST — BOF NER 2 eae ees cee a 
Setting n = k, k+1, we verify that 
mw ke Re 
Ge) 6 gelakkin) => 1) A@kina) > BGs 
j=1 ij 
holds for n = 1, n = 2; and a complete induction shows, on taking account 
of (14) §3, (p = n), that it holds for all positive integral values of n. On 
*See Chrystal: Algebra II, Ex. 26, p. 20. 
