24.8 Mr. Knight’s note respecting the demonstration 
n(n— i)(n — 2 ) A" -{-3(71 — i)(n— a) 2 . A"-!«— *)-j-(g 2 2 .«— g 
-J-w — s)A // -(«— 2 ) + (?z — 2 . n — 3 . ?z — 4+ w — 3)A // -(«— 3 ) = o 
If we take successively 4, 5, 6, &c. for we find 
<p(i+x)=A' {x-fjx‘+;f-x s -} 
+ A" {x*-f 2 x*+ -jf x s -} 
H-A'"^— -fx*+-f x s -} 
The series at bottom is L 3 (i-f a?) ; the series next above it is 
L 9 ( i-|-a?) + L 3 ( To find the upper series we have 
the same means as in the last Problem, L( i-{-a?) being a par- 
ticular value of <p(i-f-a?) ; make then A'=i, A"= — A ,,# =--; 
our equation becomes L(i-j-a?)=.r — ~~ x s — 
— f {L* ( i+x) + L 3 ( l+x) } 
+ f-L‘(i+x) 
whence a?— ~ x'— See. = L(i + .T) + ~L s (i-f-a?) 
+ T L 3 (i+a:),and 
<K i+^)=A' |L( i-j-x)-f yL 2 (i-|-i’)-}- - 6 r L 3 ( 1 -\-x ) j 
+A" |L 8 (i+^) -r L 3 ( 1+-T) } 
-{-A'" L 3 (i+.r) 
which is the complete solution of the proposed equation. 
As, however, this result has been obtained by the inspec- 
tion of only a few terms of two series, a doubt may be 
entertained with respect to its truth : make therefore w=x= 
y—z, in equation (3), it will become 4<p( i-|-.r)=6<p( 1+a:) 2 
— 4<p( 1-f-a;) 3 -{- ^1+j:) 4 , which by the substitution of the 
