83 
It now remains to prove that when r is put equal to x all 
the coefficients thus found, except Yq, will vanish, and that 
whether the sign of x is first changed or not. 
Consider the general expression 
I [n — + V){n — 2^ + 2)..(n — p - 1)} — 2p -f 3){?2 — 2p + 4),. 
(?z - p + 1) I ^ I - 2p + 5) - 2p + 6). . . - p + 3) . 
+ (n + l){n + 2)... (n + p-l) 
The difference of two consecutive terms, omitting the 
binomial coefficient is 
I (ii - 2p + 2 < 2 ' — l)(?^ -2p + 2q) — {n-p + 2q - 2){n-p + 2^' - 1) j- 
X I (?^ - 2p + 2^' + l){n -2p + 2q + 2)...(n -p + 2q - 3)| 
= - (p - 1) {2n- 3p - 2 + iq)[ (^ - 2p + 2q +1)... -p + 22^ - 3) j- 
whereby the series can be divided into two series of an 
order lower in dimensions in n by 2. 
As this is somewhat difficult to follow I will use a special 
case, though the method in the general case is similar. 
Consider 
?i-9 . n-'^ . n-1 . n-Q — b{n - 7)(n - 6)(n - 5)(n - 4) 
+ 10(?^ — 5)(n — i)(n — 3)(n — 2) — l^{n — 3)(^ — 2){n — X)n 
+ 5(71 - \)n{n + 1)(tz ■\-2)-{n^ l)(7i + 2){n + 3){n + 4) 
breaking the difference of terms up as before, this expression 
= -4(2T7-17){7i^. ^i:^--4:(n-5)(n-i) + 6(?i-3)(n-2) 
- 4(n - l)n + {n + l)(n + 2) 
-16 -177 -7 . 77 - 6 - 8(t7 - 5)(t7 - 4) + 18(?7 - 3)(t7 - 2) 
- 16(t7 - 1)t7 + 5(t7 + 1)(t7 + 2) I 
= — 4(2;7 — 13)|77 - 7. 77-6-... +(77 + 1)(t7 + 2) | 
+ 64 1 77 - 5 . 77 - 4 - 3t7 - 3 . 77 - 2 + 377 - 1 . 77 — 77 + 1 . 77 + 2 1 
The null value therefore of the original expressions will 
therefore be made to depend upon that of the expressions 
{n-2q+ 1)(t7 -2q + 2) - q{n - 2q + 3)(77 - 2q+ i) + tfec. 
+ (-1)V+1)(77+2) 
and (77 - 2(7 + 1) - g(77 - 2g + 3) + &c. + ( - 1 )'^(t7 + 1). 
Now ( 77 - 3) - 2 ( 77 - 1) + ( 77 + 1) = 0, 
