INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5 (2005), #A15
GENERALIZATION OF A BINOMIAL IDENTITY OF SIMONS
Emanuele Munarini
Dipartimento di Matematica, Politecnico di Milano, P.za Leonardo da Vinci 32, 20133 Milano, Italy
munarini@mate.polimi.it
Received: 7/30/03, Revised: 6/16/05, Accepted: 7/6/05, Published: 7/8/05
Abstract
We give a generalization of a binomial identity due to S. Simons using Cauchy’s integral
formula.
In [2] Simons proved a binomial identity which can be equivalently written as
n n n+k
k=0
k
k
k
x =
n n n+k
k=0
k
k
(−1)n−k (1 + x)k .
(1)
In [1] and [3] this identity is proved in different short ways. In this note, using Cauchy’s
integral formula as in [3], we give a generalization of (1). Consider the term
n (1 + yt)α α
β
fn = [t ]
=
xk y n−k
(1 − xt)β
n
−
k
k
k=0
n
where βk = β(β + 1) · · · (β + k − 1)/k! and α , β , x , y are indeterminates. By
Cauchy’s integral formula we have
α
1
(1 + yz)α dz
n (1 + yt)
fn = [t ]
=
.
(1 − xt)β
2πi
(1 − xz)β z n+1
With the substitution z = w/(1 − sw) , where s is an indeterminate, we have dz =
dw/(1 − sw)2 and
1
dw
(1 + (y − s)w)α
fn =
(1 − sw)β−α+n−1 n+1
β
2πi
(1 − (x + s)w)
w
α
(1 + (y − s)t)
= [tn ]
(1 − st)β−α+n−1 .
(1 − (x + s)t)β
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5 (2005), #A15
2
We now distinguish some cases. First let s = y . Then
n β−α+n−1
β−α+n−1
β
n (1 − yt)
fn = [t ]
=
(−1)n−k (x + y)k y n−k .
β
n−k
k
(1 − (x + y)t)
k=0
Hence we have the identity
n n α
β
β−α+n−1
β
k n−k
=
x y
(−1)n−k (x + y)k y n−k .
n
−
k
k
n
−
k
k
k=0
k=0
Substituting β with β + 1 identity (2) becomes
n n α
β + k k n−k β − α + n β + k
=
x y
(−1)n−k (x + y)k y n−k .
n
−
k
k
n
−
k
k
k=0
k=0
In particular for α = β identity (3) becomes
n n α
α + k k n−k n α + k
=
x y
(−1)n−k (x + y)k y n−k .
n−k
k
k
k
k=0
k=0
(2)
(3)
(4)
In particular, when α = n we have Simons’s identity (1).
Suppose now that s = −x . Then
n
α
β−α+n−1
fn = [t ](1 + (x + y)t) (1 + xt)
=
n α β−α+n−1
k=0
k
n−k
(x + y)k xn−k
and hence
n n α
β
α β−α+n−1
k n−k
=
x y
(x + y)k xn−k .
n
−
k
k
k
n
−
k
k=0
k=0
Substituting β with β + 1 , we have
n n α
β + k k n−k α β − α + n
x y
(x + y)k xn−k .
=
n−k
k
k
n−k
k=0
k=0
Finally for α = β we get
n n α
α + k k n−k n α
=
x y
(x + y)k xn−k .
n
−
k
k
k
k
k=0
k=0
Let now y = 2s and β = 2α − n + 1 . Then
α2α − 2k (1 − s2 t2 )α
n
fn = [t ]
=
(−1)k s2k (x + s)n−2k .
2α−n+1
k
n − 2k
(1 − (x + s)t)
k≥0
(5)
(6)
(7)
3
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5 (2005), #A15
Hence
n α 2α − k
k=0
k
n−k
n/2 k n−k
(2s) x
=
k=0
n
k
2α − 2k
(−1)k s2k (x + s)n−2k .
n − 2k
(8)
Similarly, for x = −2s and α = 2β + n − 1 , after the substitution of β with β + 1 ,
we have
2 2β+n+1
2β + n + 1β + k n (1 + (y − s)t )
=
s2k (y − s)n−2k
fn = [t ]
n
−
2k
k
(1 − s2 t2 )β+1
k≥0
and thus
n 2β + n + 1 β + k
k=0
n−k
k
n/2 k n−k
(−2s) y
=
k=0
2β + n + 1 β + k 2k
s (y − s)n−2k .
n − 2k
k
(9)
References
[1] R. Chapman, A curious identity revised, The Mathematical Gazette 87 (2003), 139–
141.
[2] S. Simons, A curious identity, The Mathematical Gazette 85 (2001), 296–298.
[3] H. Prodinger, A curious identity proved by Cauchy’s integral formula, The Mathematical Gazette, to appear.