576 
PROBLEMS AND SOLUTIONS. 
[383 
have a = n, /3 = — n, y = x; hence if n be a positive or negative integer, the formula 
is an identity, but if n be fractional, the condition of convergency is x > 0, that is, 
x must be positive. 
To prove the formula (1) it is only necessary to remark, that (n being a positive 
\x — ll n 
integer) the quantity ---in is a rational fraction, the numerator and denominator 
I CG “1“ Tl — JLJ 
whereof are of the same degree n, and which becomes =1 for x = oo. Hence, decom 
posing it into simple fractions, we may write 
[x-l]> 1 ~ A r 
[x + n- If + r ' x + r 
where the summation extends from r = 0 to r = n — 1 both inclusive. And we have 
A _ (Q + r)[x- l] w ) 
r l [x + n-If }*=_,. ’ 
or, observing that [x + n — l] n = [# + n — l]«-» - - 1 [pc + r) [x + r — l]' - , we have 
A= \ 1> ~ 1 3 w 1 = [-y- 1 ? 
{[a? + n — l] n_r_1 [x + r — l] r j x =-r [n — r— l] w_r_1 [— l] r 
= (-)" [n + r T = ( _ )n +r i n + r T +r = (_\n+r i n + r T +1 
[n — r— l] >l-r-1 (—) r [rf ' [n — r — l]»-»— 1 [r] r \ff ' [rf. [rf 
Hence the formula is 
[x — l] w 
[x + n — 1]” 
1 + (—)”.$»•(—) r 
[n + rf' +1 1 
[rf [rf x + r' 
or, as this may also be written, 
(s-l)(a?-2 )..{x-n) = i 1 _ n (n- - 1-) 1 n (n 2 -l 2 ) (rf - 2 2 ) 1 \ 
x (x + 1) .. (x + n — 1) ' ' ( ' x l 2 ' x + 1 l 2 .2 2 ' x + 2 ') 
which is the formula in question. 
[Vol. ir. pp. 51, 52.] 
1512. (Proposed by Professor Cayley.)—It is possible to construct a hexagon 
123456, inscribed in a conic, and such that the diagonals 14, 25, 36 pass respectively 
through the Pascalian points (intersections of opposite sides) 23, 56; 34, 61; 45, 12. 
Given the points 1, 2; 4, 5; to construct the hexagon.