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FACTORIALS, AND DERIVATIONS.
465
to obtain [to + n] 2 we have to multiply this by to + r — I ; in regard to the first term
to, I write the multiplier under the form (m — l) + n, and in regard to the second
term n, under the form m + (n — 1) ; the process then stands
[to + n] 1 = to + n
TO + 'll — 1 = (to — 1) + n TO + (n — 1)
m(m — 1) + mn
+ mn + n(n — 1)
\?n+iif = rn (to — 1 ) + 2mn + n (n — 1)
= [m] 2 + 2 [to] 1 [n] 1 + [n] 2 .
To form [m + ?i] 3 we have to multiply by m + n — 2 ; in regard to the first term, this
is written under the form (in — 2) + n ; in regard to the second term under the form
(to — 1) + (n — 1) ; and in regard to the third term under the form to + (n — 2) ; the
product is thus obtained in the form
[to] 3 + [to] 2 [w] 1
+ 2 [to] 2 [n] 1 + 2 [/n] 1 [?i] 2
+ [to] 1 [w] 2 + [n] 3
= [to] 3 + 3 [/?i] 2 [w] 1 + 3 [??i] x [n] 2 + [w] 3 ,
and so on ; the law of the terms is obvious ; and the numerical coefficients are in
effect obtained as follows :
1, 1
x by 1, 1
1, 1
1, 1
1, 2, 1
x by 1, 1
1, 2, 1
1,
1, 3, 3, Î
x by 1, 1
&c.,
viz. by precisely the same process as is used in finding the numerical coefficients in
the powers (to + îi) 1 , (m+n) 2 , (m + n) 3 , &c. ; and we thus see that for any integer value
r of the index, we have a factorial binomial theorem, wherein the numerical coefficients
are the same as in the binomial theorem for the same index.
C. VIII.
59