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

∑_{k=0}^{n} \binom{n}{k}\binom{n + k}{k}x^{k} = ∑_{k=0}^{n} \binom{n}{k}\binom{n + 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

f_{n} = [t^{n}]\frac{(1 + yt)^{α}}{(1 − xt)^{β}} = ∑_{k=0}^{n} \binom{α}{n − k}\begin{aligned}( \\ ( & {β \atop k} & ) \\ & & ) \end{aligned}x^{k}y^{n−k}

where \begin{aligned}( \\ ( & {β \atop k} & ) \\ & & ) \end{aligned} = β(β + 1) ⋯ (β + k − 1)/k! and α, β, x, y are indeterminates. By Cauchy’s integral formula we have

f_{n} = [t^{n}]\frac{(1 + yt)^{α}}{(1 − xt)^{β}} = \frac{1}{2πi} ∮ \frac{(1 + yz)^{α}}{(1 − xz)^{β}} \frac{dz}{z^{n+1}} .

With the substitution z = w/(1 − sw) , where s is an indeterminate, we have dz = dw/(1 − sw)^{2} and

f_{n} = \frac{1}{2πi} ∮ \frac{(1 + (y − s)w)^{α}}{(1 − (x + s)w)^{β}} (1 − sw)^{β−α+n−1} \frac{dw}{w^{n+1}}
= [t^{n}]\frac{(1 + (y − s)t)^{α}}{(1 − (x + s)t)^{β}} (1 − st)^{β−α+n−1} .

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5 (2005), #A15 2

We now distinguish some cases. First let s = y. Then

f_{n} = [t^{n}]\frac{(1 − yt)^{β−α+n−1}}{(1 − (x + y)t)^{β}} = ∑_{k=0}^{n} \binom{β − α + n − 1}{n − k}\begin{matrix}( \\ ( & {β \atop k} & ) \\ & & ) \end{matrix}(−1)^{n−k}(x + y)^{k}y^{n−k} .

Hence we have the identity

∑_{k=0}^{n} \binom{α}{n − k}\begin{aligned}( \\ ( & {β \atop k} & ) \\ & & ) \end{aligned}x^{k}y^{n−k} = ∑_{k=0}^{n} \binom{β − α + n − 1}{n − k}\begin{matrix}( \\ ( & {β \atop k} & ) \\ & & ) \end{matrix}(−1)^{n−k}(x + y)^{k}y^{n−k} . (2)

Substituting β with β + 1 identity (2) becomes

∑_{k=0}^{n} \binom{α}{n − k}\binom{β + k}{k}x^{k}y^{n−k} = ∑_{k=0}^{n} \binom{β − α + n}{n − k}\binom{β + k}{k}(−1)^{n−k}(x + y)^{k}y^{n−k} . (3)

In particular for α = β identity (3) becomes

∑_{k=0}^{n} \binom{α}{n − k}\binom{α + k}{k}x^{k}y^{n−k} = ∑_{k=0}^{n} \binom{n}{k}\binom{α + k}{k}(−1)^{n−k}(x + y)^{k}y^{n−k} .  (4)

In particular, when α = n we have Simons’s identity (1).

Suppose now that s = −x. Then

f_{n} = [t^{n}](1 + (x + y)t)^{α}(1 + xt)^{β−α+n−1} = ∑_{k=0}^{n} \binom{α}{k}\binom{β − α + n − 1}{n − k}(x + y)^{k}x^{n−k}

and hence

∑_{k=0}^{n} \binom{α}{n − k}\begin{aligned}( \\ ( & {β \atop k} & ) \\ & & ) \end{aligned}x^{k}y^{n−k} = ∑_{k=0}^{n} \binom{α}{k}\binom{β − α + n − 1}{n − k}(x + y)^{k}x^{n−k} .  (5)

Substituting β with β + 1 , we have

∑_{k=0}^{n} \binom{α}{n − k}\binom{β + k}{k}x^{k}y^{n−k} = ∑_{k=0}^{n} \binom{α}{k}\binom{β − α + n}{n − k}(x + y)^{k}x^{n−k} .  (6)

Finally for α = β we get

∑_{k=0}^{n} \binom{α}{n − k}\binom{α + k}{k}x^{k}y^{n−k} = ∑_{k=0}^{n} \binom{n}{k}\binom{α}{k}(x + y)^{k}x^{n−k} .  (7)

Let now y = 2s and β = 2α − n + 1 . Then

f_{n} = [t^{n}]\frac{(1 − s^{2}t^{2})^{α}}{(1 − (x + s)t)^{2α−n+1}} = ∑_{k≥0} \binom{α}{k}\binom{2α − 2k}{n − 2k}(−1)^{k}s^{2k}(x + s)^{n−2k} .

Hence

∑_{k=0}^{n} \binom{α}{k}\binom{2α − k}{n − k}(2s)^{k}x^{n−k} = ∑_{k=0}^{⌊n/2⌋} \binom{n}{k}\binom{2α − 2k}{n − 2k}(−1)^{k}s^{2k}(x + s)^{n−2k} . (8)

Similarly, for x = −2s and α = 2β + n − 1 , after the substitution of β with β + 1 , we have

f_{n} = [t^{n}]\frac{(1 + (y − s)t^{2})^{2β+n+1}}{(1 − s^{2}t^{2})^{β+1}} = ∑_{k≥0} \binom{2β + n + 1}{n − 2k}\binom{β + k}{k}s^{2k}(y − s)^{n−2k}

and thus

∑_{k=0}^{n} \binom{2β + n + 1}{n − k}\binom{β + k}{k}(−2s)^{k}y^{n−k} = ∑_{k=0}^{⌊n/2⌋} \binom{2β + n + 1}{n − 2k}\binom{β + k}{k}s^{2k}(y − s)^{n−2k} .(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.