Open AccessArticle

Combinatorial Identities Concerning Binomial Quotients

Yulei Chen

¹ and

Dongwei Guo

^2,*

School of Mathematics and Statistics, Zhoukou Normal University, Zhoukou 466001, China

School of Economics and Management, Nanjing University of Science and Technology, Nanjing 210094, China

Author to whom correspondence should be addressed.

Symmetry 2024, 16(6), 746; https://doi.org/10.3390/sym16060746

Submission received: 16 April 2024 / Revised: 31 May 2024 / Accepted: 13 June 2024 / Published: 14 June 2024

(This article belongs to the Section Mathematics)

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Abstract

Making use of a telescoping approach, three types of sums of binomial quotients are examined. The summation terms of the two types of alternating sums have symmetry (i.e., their numerators and denominators are completely symmetric). We obtained a series of their explicit sums. Furthermore, by means of binomial relations, three recurrence relations of the sums are derived. In addition, series of double summation formulae involving binomial quotients are established.

Keywords:

binomial quotient; telescoping; combinatorial identities

1. Introduction and Motivation

Combinatorial identities have widespread applications in many fields, such as mathematics, computer sciences and physics. In [1], Gould recorded more than 500 binomial identities. Among many others, he listed some interesting summation formulae of binomial quotients, such as [1] Equations (4.22), (4.26), (4.23) and (4.27):

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{m}{k})}{(\binom{2 m}{2 k})} = \frac{1 + {(- 1)}^{m}}{2} \cdot \frac{2 m + 1}{m + 1}; \end{matrix}

(1)

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{2 m}{2 k})}{(\binom{m}{k})} = \frac{1 + {(- 1)}^{m}}{2} \cdot \frac{1}{1 - m}; \end{matrix}

(2)

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{m}{k})}{(\binom{2 m + 1}{2 k + 1})} = \frac{1 - {(- 1)}^{m}}{2} \cdot \frac{1}{m + 2} + {(- 1)}^{m}; \end{matrix}

(3)

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{2 m + 1}{2 k + 1})}{(\binom{m}{k})} = \frac{1 - {(- 1)}^{m}}{2 m} + 1 . \end{matrix}

(4)

For more combinatorial identities with inverse binomial coefficients, the reader can refer to [2,3,4,5,6,7,8]. In particular, the following results are well known:

\begin{matrix} S_{m} : = \sum_{k = 0}^{m} {(\binom{m}{k})}^{- 1} = \frac{m + 1}{2^{m}} \sum_{k = 0}^{m} \frac{2^{k}}{k + 1} . \end{matrix}

(5)

Furthermore,

S_{m}

satisfies the recurrence relation

\begin{matrix} S_{m} = 1 + \frac{m + 1}{2 m} S_{m - 1} . \end{matrix}

(6)

The above results are the subject of problem 1 in the afternoon session of the 1958 Putnam Exam. They were recorded by Comtet [9] (Exercise 15, p. 294) and Graham et al. [10] (Exercise 5.100, pp. 542–543). There are different proofs for the results, which can be found in [2,3,4,5]. In addition, Mansour [2] also obtained the following weighted result:

\sum_{k = 0}^{m} \frac{k}{(\binom{m}{k})} = \frac{(m + 1) (2^{m} - 1)}{2^{m}} + \frac{1}{2^{m + 1}} \sum_{k = 0}^{m - 2} \frac{(m - k) (m - k - 1) 2^{k}}{k + 1} .

Among many methods, such as the generating function method [11,12,13], WZ method [14,15], Riordan array approach [16,17] and hypergeometric series technique [18], to prove combinatorial identities, the telescoping technique proposed by Zeilberger [19,20] is a quite efficient approach to computing binomial sums. This is the main method used in this paper, and thus we will give a brief introduction to it here.

For summing

\sum_{k = ℓ}^{m} a_{k}

, we construct a new sequence

A_{k}

such that

a_{k} = A_{k + 1} - A_{k}

, and then

\begin{matrix} \sum_{k = ℓ}^{m} a_{k} = \sum_{k = 1}^{m} (A_{k + 1} - A_{k}) = A_{m + 1} - A_{ℓ} . \end{matrix}

Recently, Chu and Guo [21] examined, by means of telescoping [22] and the linearization method [23], the following alternating sums of binomial quotients:

\begin{matrix} Q_{λ, ε} (m) : = \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{m + λ}{2 k + ε})}{(\binom{m}{k})}, \end{matrix}

(7)

where

λ, m \in N_{0}

and

ε \in {0, 1}

Motivated by Equations (1)–(4) and (7) above, in this paper, we shall, by means of telescoping, calculate the following sums of binomial quotients:

Ω_{m} (τ, δ, ε) : = \sum_{k = 0}^{m} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m + ε}{k})}, with τ, ε \in N_{0} .

\begin{matrix} Φ_{m} (τ, δ, ε) : = \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + δ}{k + ε})}, with δ \geq ε . \end{matrix}

Ψ_{m} (τ, δ, ε) : = \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{m + δ}{k + ε})}{(\binom{τ m - k}{m - k})}, with δ \leq ε .

As we can see, the summation terms of

Φ_{m} (τ, δ, ε)

and

Ψ_{m} (τ, δ, ε)

have symmetry (i.e., their numerators and denominators are completely symmetrical).

Furthermore, we shall further explore the following double finite series:

\begin{matrix} P_{λ, ε} (m) : = \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + λ}{2 k})}{(\binom{m}{j})}, with λ \leq m, ε = 0, 1, \end{matrix}

whose results are related to

Q_{λ, ε} (m)

For an indeterminate x, the symbol

{(x)}_{n}

will be used in the next sections, with the rising factorial defined by quotients of the gamma function

{(x)}_{n} = \frac{Γ (x + n)}{Γ (x)} = \{\begin{matrix} 1, & n = 0, \\ x (x + 1) (x + 2) \dots (x + n - 1), & n = 1, 2, \dots . \end{matrix}

The generalized hypergeometric series used by Bailey [24] are defined as follows:

\begin{matrix} {}_{1 + p}F_{q} [\begin{matrix} a_{0}, a_{1}, \dots, a_{p} \\ b_{1}, \dots, b_{q} \end{matrix} | z] = \sum_{k = 0}^{\infty} \frac{{(a_{0})}_{k} {(a_{1})}_{k} \dots {(a_{p})}_{k}}{k! {(b_{1})}_{k} \dots {(b_{q})}_{k}} . \end{matrix}

Then, the above three sums can be rewritten as hypergeometric series

{}_{3}F_{2}

\begin{matrix} Ω_{m} (τ, δ, ε) = (\binom{τ m + δ}{m}) {}_{3}F_{2} [\begin{matrix} - m, & 1, & 1 \\ - m - ε, & (τ - 1) m + δ + 1 \end{matrix} | 1], \\ Φ_{m} (τ, δ, ε) = \frac{(\binom{τ m}{m})}{(\binom{m + δ}{ε})} {}_{3}F_{2} [\begin{matrix} - m, & ε + 1, & 1 \\ - τ m, & ε - δ - m \end{matrix} | 1], \\ Ψ_{m} (τ, δ, ε) = \frac{(\binom{m + δ}{ε})}{(\binom{τ m}{m})} {}_{3}F_{2} [\begin{matrix} - τ m, & ε - δ - m, & 1 \\ - m, & ε + 1 \end{matrix} | 1] . \end{matrix}

In particular, when setting

τ = 1, δ - ε = 2

, the sum

Ω_{m} (1, δ, ε)

can be evaluated with Saalschütz’s theorem. Furthermore, these series are difficult to calculate using the classical Saalschütz, Dixon, Watson and Whipple theorems. Many scholars have studied the hypergeometric series concerning

{}_{3}F_{2}

, such as Chu [23], Chen [25], Li and Chu [26] and Milgram [27]. However, the series examined in this paper are also difficult to calculate directly from their results.

2. Summation Formulae $Ω_{m} (τ, δ, ε)$

In this section, we shall evaluate

Ω_{m} (τ, δ, ε)

, defined by the binomial quotient

Ω_{m} (τ, δ, ε) : = \sum_{k = 0}^{m} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m + ε}{k})}, with τ, ε \in N_{0} .

Theorem 1.

Ω_{m} (τ, δ, 0)

\begin{matrix} \sum_{k = 0}^{m} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m}{k})} = \frac{(τ - 1) m + δ}{(τ - 1) m + δ - 1} (\binom{τ m + δ}{m}) - \frac{m + 1}{(τ - 1) m + δ - 1} . \end{matrix}

(8)

Proof.

Notice that the quotient can be rewritten as

\frac{(\binom{τ m + δ}{m - k})}{(\binom{m}{k})} = \frac{τ m + δ}{τ m - m + δ - 1} (A_{k + 1} - A_{k}) - \frac{τ m + δ}{τ m - m + δ - 1} (A_{k + 1} - A_{k}),

where

A_{k} = \frac{(\binom{τ m + δ - 1}{m - k - 1})}{(\binom{m}{k})} and A_{k} = \frac{(\binom{τ m + δ}{m - k})}{(\binom{m}{k})} .

By means of the telescoping technique, the sum becomes

\begin{matrix} \sum_{k = 0}^{m} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m}{k})} & = 1 + \frac{τ m + δ}{m} + \sum_{k = 0}^{m - 2} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m}{k})} \\ = 1 + \frac{τ m + δ}{m} + \frac{τ m + δ}{τ m - m + δ - 1} (A_{m - 1} - A_{0} - A_{m - 1} + A_{0}) . \end{matrix}

The proof follows from the fact that

A_{m - 1} = \frac{1}{m}, A_{0} = (\binom{τ m + δ - 1}{m - 1}) and A_{m - 1} = \frac{τ m + δ}{m}, A_{0} = (\binom{τ m + δ}{m}) .

□

Specially, we have the following particular cases from Theorem 1.

Corollary 1.

\begin{matrix} Ω_{m} (1, 0, 0) & = \sum_{k = 0}^{m} 1 = m + 1; \\ Ω_{m} (2, 1, 0) & = \sum_{k = 0}^{m} \frac{(\binom{2 m + 1}{m - k})}{(\binom{m}{k})} = \frac{m + 1}{m} \{(\binom{2 m + 1}{m}) - 1\} . \end{matrix}

Theorem 2.

Ω_{m} (τ, δ, 1)

\begin{matrix} \sum_{k = 0}^{m} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m + 1}{k})} & = \frac{m + 2}{{(m (τ - 1) + δ - 2)}_{2}} \\ + \frac{(τ - 1) m^{2} + m (τ + δ - 3) + δ - 3}{{(m (τ - 1) + δ - 2)}_{2}} (\binom{τ m + δ}{m + 1}) . \end{matrix}

(9)

Proof.

Write the summand as

\begin{matrix} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m + 1}{k})} & = \frac{m + 2}{(τ - 1) m + δ - 2} (B_{k + 1} - B_{k}) \\ - \frac{(τ - 1) m^{2} + m (τ + 2 δ - 4) + 2 δ - 4}{{(m (τ - 1) + δ - 2)}_{2}} (B_{k + 1} - B_{k}), \end{matrix}

where

B_{k} = \frac{(\binom{τ m + δ}{m - k + 1})}{(\binom{m + 2}{k + 1})} and B_{k} = \frac{(\binom{τ m + δ}{m - k + 1})}{(\binom{m + 1}{k})} .

By using telescoping, we can evaluate the sum

\begin{matrix} \sum_{k = 0}^{m} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m + 1}{k})} & = \frac{m + 2}{(τ - 1) m + δ - 2} (B_{m + 1} - B_{0}) \\ - \frac{(τ - 1) m^{2} + m (τ + 2 δ - 4) + 2 δ - 4}{{(m (τ - 1) + δ - 2)}_{2}} (B_{m + 1} - B_{0}) . \end{matrix}

Then, the proof follows from the fact that

B_{m + 1} = B_{m + 1} = 1

and

B_{0} = \frac{1}{m + 2} (\binom{τ m + δ}{m + 1}) and B_{0} = (\binom{τ m + δ}{m + 1}) .

□

For spacial cases, we obtain from Theorem 2 the corollary below.

Corollary 2.

\begin{matrix} Ω_{m} (2, 3, 1) & = \sum_{k = 0}^{m} \frac{(\binom{2 m + 3}{m - k})}{(\binom{m + 1}{k})} = \frac{1}{m + 1} + \frac{m}{m + 1} (\binom{2 m + 3}{m + 1}); \\ Ω_{m} (3, 3, 1) & = \sum_{k = 0}^{m} \frac{(\binom{3 m + 3}{m - k})}{(\binom{m + 1}{k})} = \frac{m + 2}{{(2 m + 1)}_{2}} + \frac{2 m^{2} + 3 m}{{(2 m + 1)}_{2}} (\binom{3 m + 3}{m + 1}); \\ Ω_{m} (4, 3, 1) & = \sum_{k = 0}^{m} \frac{(\binom{4 m + 3}{m - k})}{(\binom{m + 1}{k})} = \frac{m + 2}{{(3 m + 1)}_{2}} + \frac{3 m^{2} + 4 m}{{(3 m + 1)}_{2}} (\binom{4 m + 3}{m + 1}) . \end{matrix}

Theorem 3.

The recurrence relation is expressed as

\begin{matrix} Ω_{m} (τ, δ, ε) = \frac{m + ε + 1}{m + ε} Ω_{m} (τ, δ, ε - 1) - Ω_{m + 1} (τ, δ - τ, ε - 1) + (\binom{τ m + δ}{m + 1}) . \end{matrix}

(10)

Proof.

By means of the relation

{(\binom{m + ε}{k})}^{- 1} = {(\binom{m + ε - 1}{k})}^{- 1} - \frac{k}{m + ε} {(\binom{m + ε - 1}{k})}^{- 1},

the sum can then be rewritten as

\begin{matrix} \sum_{k = 0}^{m} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m + ε}{k})} = \frac{m + ε + 1}{m + ε} \sum_{k = 0}^{m} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m + ε - 1}{k})} - \sum_{k = 0}^{m} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m + ε}{k + 1})} . \end{matrix}

By replacing k with

k - 1

, the second sum on the right-hand side becomes

V_{m + 1} : = \sum_{k = 0}^{m} \frac{(\binom{τ m + δ}{m - k})}{(\binom{m + ε}{k + 1})} = \sum_{k = 1}^{m + 1} \frac{(\binom{τ m + δ}{m - k + 1})}{(\binom{m + ε}{k})} .

Thus, we have

V_{m} = \sum_{k = 1}^{m} \frac{(\binom{τ m - τ + δ}{m - k})}{(\binom{m + ε - 1}{k})} = Ω_{m} (τ, δ - τ, ε - 1) - (\binom{τ m - τ + δ}{m}),

which confirms the recurrence relation stated in Theorem 3. □

It is not difficult to find the corollary below by means of Theorems 2 and 3.

Corollary 3.

\begin{matrix} Ω_{m} (2, 3, 2) & = \sum_{k = 0}^{m} \frac{(\binom{2 m + 3}{m - k})}{(\binom{m + 2}{k})} = \frac{m^{3} + 2 m^{2} - m + 2}{{(m)}_{3}} (\binom{2 m + 3}{m + 1}) - \frac{2 (m + 3)}{{(m)}_{3}}; \\ Ω_{m} (2, 3, 3) & = \sum_{k = 0}^{m} \frac{(\binom{2 m + 3}{m - k})}{(\binom{m + 3}{k})} = \frac{m^{2} (m^{2} - 1) (m + 4) + 12 (m - 2)}{{(m - 1)}_{4} (m - 1)} (\binom{2 m + 3}{m + 1}) + \frac{6 (m + 4)}{{(m - 1)}_{4}} . \end{matrix}

3. Summation Formulae $Φ_{m} (τ, δ, ε)$

In this section, we shall deal with the sum

Φ_{m} (τ, δ, ε)

defined by the alternating binomial quotient:

\begin{matrix} Φ_{m} (τ, δ, ε) : = \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + δ}{k + ε})}, with δ \geq ε . \end{matrix}

Theorem 4.

Φ_{m} (τ, 1, 1)

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 1}{k + 1})} = {(- 1)}^{m} \frac{m + 2}{τ m + 3} + \frac{1}{τ m + 3} (\binom{τ m + 1}{m + 1}) . \end{matrix}

(11)

Proof.

We write the summand as

\begin{matrix} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 1}{k + 1})} = \frac{τ m - m + 1}{τ m + 3} (C_{k} + C_{k + 1}), where C_{k} = \frac{(\binom{τ m - k + 1}{m - k})}{(\binom{m + 1}{k + 1})} . \end{matrix}

By using telescoping, we have

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 1}{k + 1})} & = {(- 1)}^{m} + \sum_{k = 0}^{m - 1} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 1}{k + 1})} \\ = {(- 1)}^{m} + \frac{τ m - m + 1}{τ m + 3} {C_{0} - {(- 1)}^{m} C_{m}} . \end{matrix}

Then the proof follows from the fact that

C_{0} = \frac{1}{m + 1} (\binom{τ m + 1}{m}) and C_{m} = 1 .

□

Theorem 5.

Φ_{m} (τ, 1, 0)

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 1}{k})} = \frac{(m + 1) (τ m + 3) + 1}{{(τ m + 2)}_{2}} (\binom{τ m + 1}{m + 1}) + {(- 1)}^{m} \frac{m + 2}{{(τ m + 2)}_{2}} . \end{matrix}

(12)

Proof.

We write the summand as

\begin{matrix} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 1}{k})} = \frac{m + 2}{τ m + 2} (D_{k} + D_{k + 1}) - \frac{m + 2}{τ m + 3} (D_{k} + D_{k + 1}), \end{matrix}

where

\begin{matrix} D_{k} = \frac{(\binom{τ m - k + 1}{m - k + 1})}{(\binom{m + 1}{k})} and D_{k} = \frac{(\binom{τ m - k + 1}{m - k + 1})}{(\binom{m + 2}{k + 1})} . \end{matrix}

By using telescoping, we can evaluate the sum

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 1}{k})} = \frac{m + 2}{τ m + 2} {D_{0} + {(- 1)}^{m} D_{m + 1}} - \frac{m + 2}{τ m + 3} {D_{0} + {(- 1)}^{m} D_{m + 1}} . \end{matrix}

Then, the proof follows from the fact that

D_{m + 1} = D_{m + 1} = 1

and

D_{0} = (\binom{τ m + 1}{m + 1}) and D_{0} = \frac{1}{m + 2} (\binom{τ m + 1}{m + 1}) .

□

By setting

τ = 2

, we obtain from Theorems 4 and 5 the following corollary.

Corollary 4.

\begin{matrix} Φ_{m} (2, 1, 1) & = \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{2 m - k}{m - k})}{(\binom{m + 1}{k + 1})} = {(- 1)}^{m} \frac{m + 2}{2 m + 3} + \frac{1}{2 m + 3} (\binom{2 m + 1}{m + 1}); \\ Φ_{m} (2, 1, 0) & = \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{2 m - k}{m - k})}{(\binom{m + 1}{k})} = \frac{(m + 1) (2 m + 3) + 1}{{(2 m + 2)}_{2}} (\binom{2 m + 1}{m + 1}) + {(- 1)}^{m} \frac{m + 2}{{(2 m + 2)}_{2}} . \end{matrix}

Theorem 6.

Φ_{m} (τ, 2, 2)

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 2}{k + 2})} = \frac{2}{(m + 2) (τ m + 4)} (\binom{τ m + 1}{m + 1}) + {(- 1)}^{m} \frac{m + 3}{τ m + 4} . \end{matrix}

(13)

Proof.

By keeping in mind the summation term

\frac{(\binom{τ m - k}{m - k})}{(\binom{m + 2}{k + 2})} = (E_{k + 1} + E_{k}) - \frac{m + 3}{τ m + 4} (E_{k + 1} + E_{k}),

where

E_{k} = \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 2}{k + 2})} and E_{k} = \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 3}{k + 3})},

then by means of telescoping, we have

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 2}{k + 2})} & = {(- 1)}^{m} + \sum_{k = 0}^{m - 1} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 2}{k + 2})} \\ = {(- 1)}^{m} + E_{0} - {(- 1)}^{m} E_{m} - \frac{m + 3}{τ m + 4} (E_{0} - {(- 1)}^{m} E_{m}) . \end{matrix}

By substituting the values

E_{m} = E_{m} = 1

and

E_{0} = \frac{(\binom{τ m}{m})}{(\binom{m + 2}{2})} and E_{0} = \frac{(\binom{τ m}{m})}{(\binom{m + 3}{3})}

into the above equation, we can then find the desired identity. □

Theorem 7.

Φ_{m} (τ, 2, 1)

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 2}{k + 1})} = \frac{τ m^{2} + τ m + 4 m + 6}{(m + 2) {(τ m + 3)}_{2}} (\binom{τ m + 1}{m + 1}) + {(- 1)}^{m} \frac{m + 3}{{(τ m + 3)}_{2}} . \end{matrix}

(14)

Proof.

We rewrite the summand as

\frac{(\binom{τ m - k}{m - k})}{(\binom{m + 2}{k + 1})} = \frac{m + 3}{τ m + 3} (F_{k + 1} + F_{k}) - \frac{m + 3}{τ m + 4} (F_{k + 1} + F_{k}),

where

F_{k} = \frac{(\binom{τ m - k + 1}{m - k + 1})}{(\binom{m + 2}{k + 1})} and F_{k} = \frac{(\binom{τ m - k + 1}{m - k + 1})}{(\binom{m + 3}{k + 2})},

Then, by making use of the telescoping approach, we obtain

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 2}{k + 1})} = \frac{m + 3}{τ m + 3} (F_{0} + {(- 1)}^{m} F_{m + 1}) - \frac{m + 3}{τ m + 4} (F_{0} + {(- 1)}^{m} F_{m + 1}) . \end{matrix}

The proof follows from the fact that

F_{m + 1} = F_{m + 1} = 1

and

F_{0} = \frac{1}{m + 2} (\binom{τ m + 1}{m + 1}) and F_{0} = \frac{2}{(m + 2) (m + 3)} (\binom{τ m + 1}{m + 1}) .

□

For spacial cases, when

τ = 2

, we obtain from Theorems 6 and 7 the corollary below.

Corollary 5.

\begin{matrix} Φ_{m} (2, 2, 2) & = \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{2 m - k}{m - k})}{(\binom{m + 2}{k + 2})} = \frac{1}{{(m + 2)}^{2}} (\binom{2 m + 1}{m + 1}) + {(- 1)}^{m} \frac{m + 3}{2 m + 4}; \\ Φ_{m} (2, 2, 1) & = \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{2 m - k}{m - k})}{(\binom{m + 2}{k + 1})} = \frac{m^{2} + 3 m + 3}{{(m + 2)}^{2} (2 m + 3)} (\binom{2 m + 1}{m + 1}) + {(- 1)}^{m} \frac{m + 3}{{(2 m + 3)}_{2}} . \end{matrix}

Theorem 8.

The recurrence relation is expressed as

\begin{matrix} Φ_{m} (τ, δ, ε) = \frac{m + δ + 1}{m + δ} Φ_{m} (τ, δ - 1, ε - 1) - Φ_{m} (τ, δ, ε - 1) . \end{matrix}

(15)

Proof.

By means of the relation

{(\binom{m + δ}{k + ε})}^{- 1} = {(\binom{m + δ - 1}{k + ε})}^{- 1} - \frac{k + ε}{m + δ} {(\binom{m + δ - 1}{k + ε})}^{- 1},

we have

\begin{matrix} Φ_{m} (τ, δ, ε) & : = \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + δ}{k + ε})} \\ = \frac{m + δ + 1}{m + δ} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + δ - 1}{k + ε})} - \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + δ}{k + ε + 1})} \\ = \frac{m + δ + 1}{m + δ} Φ_{m} (τ, δ - 1, ε) - Φ_{m} (τ, δ, ε + 1) . \end{matrix}

Then, the proof follows by making the replacement

ε \to ε - 1

. □

By means of Theorem 8 above, we obtain from Theorem 5 and Theorem 7 the following corollary.

Corollary 6

(

Φ_{m} (τ, 2, 0)

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m - k}{m - k})}{(\binom{m + 2}{k})} & = \frac{(m + 1) (τ m + 4) [(m + 2) (τ m + 3) + 2] + 4}{(m + 2) {(τ m + 2)}_{3}} (\binom{τ m + 1}{m + 1}) \\ + {(- 1)}^{m} \frac{2 (m + 3)}{{(τ m + 2)}_{3}} . \end{matrix}

4. Summation Formulae $Ψ_{m} (τ, δ, ε)$

In this section, we shall compute the following alternating sum:

Ψ_{m} (τ, δ, ε) : = \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{m + δ}{k + ε})}{(\binom{τ m - k}{m - k})}, with δ \leq ε .

Theorem 9.

Ψ_{m} (τ, 1, 1)

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{m + 1}{k + 1})}{(\binom{τ m - k}{m - k})} = \frac{1}{(\binom{τ m + 1}{m + 1})} . \end{matrix}

(16)

Proof.

Keep in mind that the summand can be rewritten as

\frac{(\binom{m + 1}{k + 1})}{(\binom{τ m - k}{m - k})} = \frac{m + 1}{τ m + 1} (G_{k} + G_{k + 1}), where G_{k} = \frac{(\binom{m}{k})}{(\binom{τ m - k}{m - k})} .

Using telescoping approach, the sum can be evaluated:

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{m + 1}{k + 1})}{(\binom{τ m - k}{m - k})} = \frac{m + 1}{τ m + 1} {G_{0} + {(- 1)}^{m} G_{m + 1}} . \end{matrix}

Then, the proof follows from the fact that

G_{m + 1} = 0

and

G_{0} = {(\binom{τ m}{m})}^{- 1}

. □

Theorem 10.

Ψ_{m} (τ, 0, 1)

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{m}{k + 1})}{(\binom{τ m - k}{m - k})} = \frac{1}{(\binom{τ m + 1}{m + 1})} - {(- 1)}^{m} \frac{τ - 1}{τ (τ m + 1)} . \end{matrix}

(17)

Proof.

Notice that the summand can be rewritten as

\frac{(\binom{m}{k + 1})}{(\binom{τ m - k}{m - k})} = \frac{m + 1}{τ m + 1} (H_{k} + H_{k + 1}) - \frac{1}{τ} (H_{k} + H_{k + 1}),

where

H_{k} = \frac{(\binom{m}{k})}{(\binom{τ m - k}{m - k})} and H_{k} = \frac{(\binom{m - 1}{k - 1})}{(\binom{τ m - k}{m - k})} .

By means of telescoping, we obtain

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{m}{k + 1})}{(\binom{τ m - k}{m - k})} & = \frac{m}{(\binom{τ m}{m})} + \sum_{k = 1}^{m - 1} {(- 1)}^{k} \frac{(\binom{m}{k + 1})}{(\binom{τ m - k}{m - k})} \\ = \frac{m}{(\binom{τ m}{m})} + \frac{m + 1}{τ m + 1} {H_{1} - {(- 1)}^{m} H_{m}} - \frac{1}{τ} {H_{1} - {(- 1)}^{m} H_{m}} . \end{matrix}

Then, the proof can be completed with the fact that

H_{m} = H_{m} = 1

and

H_{1} = \frac{1}{(\binom{τ m - 1}{m - 1})} and H_{1} = \frac{m}{(\binom{τ m - 1}{m - 1})} .

□

Theorem 11.

The recurrence relation is expressed as

\begin{matrix} Ψ_{m} (τ, δ, ε) = Ψ_{m} (τ, δ - 1, ε - 1) + Ψ_{m} (τ, δ - 1, ε) . \end{matrix}

(18)

Proof.

The proof follows by using the relation

(\binom{m + δ}{k + ε}) = (\binom{m + δ - 1}{k + ε - 1}) + (\binom{m + δ - 1}{k + ε}) .

□

5. Double Summation Formulae $P_{λ, ε} (m)$

In this section, we shall explore the following double summation formulae:

\begin{matrix} P_{λ, ε} (m) : = \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + λ}{2 k})}{(\binom{m}{j})}, with λ \leq m, ε = 0, 1 . \end{matrix}

First, we prove the following lemma.

Lemma 1.

For integers

m, λ

and j, the sequence

\begin{matrix} S_{m, λ} (x) : = \sum_{j = k}^{m} \frac{x^{j}}{(\binom{m + λ}{j})} \end{matrix}

satisfies the recurrence relation

\begin{matrix} (1 + x) S_{m + 1, λ} (x) = x \frac{m + λ + 2}{m + λ + 1} S_{m, λ} (x) + \frac{x^{k}}{(\binom{m + λ + 1}{k})} + \frac{x^{m + 2}}{(\binom{m + λ + 1}{m + 1})} . \end{matrix}

(19)

Proof.

By making use of the binomial recurrence relation

\frac{1}{(\binom{m + λ + 1}{j})} = \frac{1}{(\binom{m + λ}{j - 1})} - \frac{m + λ + 1 - j}{m + λ + 2 - j} \frac{1}{(\binom{m + λ + 1}{j - 1})},

we can manipulate the sum

\begin{matrix} S_{m + 1, λ} (x) & = \sum_{j = k}^{m + 1} \frac{x^{j}}{(\binom{m + 1 + λ}{j})} = \sum_{j = k}^{m + 1} x^{j} \{\frac{1}{(\binom{m + λ}{j - 1})} - \frac{m + λ + 1 - j}{m + λ + 2 - j} \frac{1}{(\binom{m + λ + 1}{j - 1})}\} \\ = \sum_{j = k}^{m + 1} \frac{x^{j}}{(\binom{m + λ}{j - 1})} - \sum_{j = k}^{m + 1} (1 - \frac{1}{m + λ + 2 - j}) \frac{x^{j}}{(\binom{m + λ + 1}{j - 1})} . \end{matrix}

In light of the binomial relation

(m + λ + 2 - j) (\binom{m + λ + 1}{j - 1}) = (m + λ + 1) (\binom{m + λ}{j - 1}),

we obtain that

\begin{matrix} S_{m + 1, λ} (x) = \frac{m + λ + 2}{m + λ + 1} \sum_{j = k}^{m + 1} \frac{x^{j}}{(\binom{m + λ}{j - 1})} - \sum_{j = k}^{m + 1} \frac{x^{j}}{(\binom{m + 1 + λ}{j - 1})} . \end{matrix}

Then, by taking the replacement of

j \to j + 1

, we have

\begin{matrix} S_{m + 1, λ} (x) & = \frac{m + λ + 2}{m + λ + 1} \sum_{j = k - 1}^{m + 1} \frac{x^{j + 1}}{(\binom{m + λ}{j})} - \sum_{j = k - 1}^{m + 1} \frac{x^{j + 1}}{(\binom{m + 1 + λ}{j})} \\ = \frac{m + λ + 2}{m + λ + 1} \frac{x^{k}}{(\binom{m + λ}{k - 1})} + x \frac{m + λ + 2}{m + λ + 1} S_{m, λ} (x) - \frac{x^{k}}{(\binom{m + λ + 1}{k - 1})} + \frac{x^{m + 2}}{(\binom{m + λ + 1}{m + 1})} - x S_{m + 1, λ} (x) . \end{matrix}

The proof can be completed by using

\begin{matrix} \frac{m + λ + 2}{m + λ + 1} \frac{x^{k}}{(\binom{m + λ}{k - 1})} - \frac{x^{k}}{(\binom{m + λ + 1}{k - 1})} & = \frac{m + λ + 2}{k} \frac{x^{k}}{(\binom{m + λ + 1}{k})} - \frac{m + λ + 2 - k}{k} \frac{x^{k}}{(\binom{m + λ + 1}{k})} \\ = \frac{x^{k}}{(\binom{m + λ + 1}{k})} . \end{matrix}

□

Corollary 7.

For two integers m and k, it holds that

\begin{matrix} \sum_{j = k}^{m} {(- 1)}^{j} \frac{1}{(\binom{m}{j})} = \frac{m + 1}{m + 2} \{{(- 1)}^{m} + \frac{{(- 1)}^{k}}{(\binom{m + 1}{k})}\} . \end{matrix}

(20)

Proof.

By letting

x = - 1

and

λ = 0

in the recurrence relation in Equation (19), we obtain that

0 = \frac{{(- 1)}^{k}}{(\binom{m + 1}{k})} + {(- 1)}^{m} - \frac{m + 2}{m + 1} S_{m, 0} (- 1),

which yields the desired result

S_{m, 0} (- 1) = \sum_{j = k}^{m} {(- 1)}^{j} \frac{1}{(\binom{m}{j})} = \frac{m + 1}{m + 2} \{{(- 1)}^{m} + \frac{{(- 1)}^{k}}{(\binom{m + 1}{k})}\} .

□

Theorem 12.

For integers

λ \leq m

and

ε \in 0, 1

, it holds that

\begin{matrix} \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + λ}{2 k + ε})}{(\binom{m}{j})} = 2^{λ - 1} {(- 2)}^{m} \frac{m + 1}{m + 2} + \frac{m + 1}{m + 2} Ω_{λ - 1, ε} (m + 1) . \end{matrix}

(21)

Proof.

Exchanging the order of summation and applying Corollary 7 yields

\begin{matrix} \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + λ}{2 k + ε})}{(\binom{m}{j})} & = \sum_{k = 0}^{m} (\binom{m + λ}{2 k + ε}) \sum_{k = 0}^{j} \frac{{(- 1)}^{j}}{(\binom{m}{j})} \\ = {(- 1)}^{m} \frac{m + 1}{m + 2} \sum_{k = 0}^{m} (\binom{m + λ}{2 k + ε}) + \frac{m + 1}{m + 2} Ω_{λ - 1, ε} (m + 1) . \end{matrix}

Then, the proof can be completed as follows:

\sum_{k = 0}^{m} (\binom{m + λ}{2 k + ε}) = 2^{m + λ - 1}, λ \leq m, ε = 0, 1 .

□

By employing the results of

Ω_{λ, ε} (m)

in [21], we can obtain series of finite double summation formulae on

P_{λ, ε} (m)

. For instance, if we let

λ = 4, 3, 2, 1

, then we have the following corollaries.

Corollary 8.

For

ε = 0

, we have

\begin{matrix} \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + 4}{2 k})}{(\binom{m}{j})} & = \frac{m + 1}{m + 2} \{8 {(- 2)}^{m} - 1\} + \frac{m^{2} - m + 4}{m + 2}; \\ \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + 3}{2 k})}{(\binom{m}{j})} & = \frac{m + 1}{m + 2} \{4 {(- 2)}^{m} - 1\} + \frac{m - 1}{m + 2}; \\ \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + 2}{2 k})}{(\binom{m}{j})} & = \frac{m + 1}{m + 2} \{2 {(- 2)}^{m} + 1\} - \frac{1}{2} \sum_{k = 0}^{m} \frac{1}{(\binom{m}{k})}; \\ \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + 1}{2 k})}{(\binom{m}{j})} & = \frac{m + 1}{m + 2} \{{(- 2)}^{m} + \frac{m + 3}{2}\} - \frac{m + 1}{4} \sum_{k = 0}^{m} \frac{1}{(\binom{m}{k})} . \end{matrix}

Corollary 9.

For

ε = 1

, we have

\begin{matrix} \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + 4}{2 k + 1})}{(\binom{m}{j})} & = \frac{m + 1}{m + 2} \{8 {(- 2)}^{m} + 2\} - \frac{2}{m + 2}; \\ \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + 3}{2 k + 1})}{(\binom{m}{j})} & = \frac{m + 1}{m + 2} \{4 {(- 2)}^{m} + 2\}; \\ \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + 2}{2 k + 1})}{(\binom{m}{j})} & = \frac{m + 1}{m + 2} \{2 {(- 2)}^{m} + 1\} + \frac{1}{2} \sum_{k = 0}^{m} \frac{1}{(\binom{m}{k})}; \\ \sum_{j = 0}^{m} \sum_{k = 0}^{j} {(- 1)}^{j} \frac{(\binom{m + 1}{2 k + 1})}{(\binom{m}{j})} & = \frac{m + 1}{m + 2} \{{(- 2)}^{m} - \frac{m + 1}{2}\} + \frac{m + 3}{4} \sum_{k = 0}^{m} \frac{1}{(\binom{m}{k})} . \end{matrix}

6. Concluding Comments

The telescoping approach is quite useful for establishing binomial sums. Apart from the formulae derived in the previous sections, more formulae can be found. For instance, by using the relation

\frac{(\binom{τ m + δ}{k})}{(\binom{m + k}{k})} = \frac{m}{τ m + m + δ} (I_{k + 1} + I_{k}), with I_{k} = \frac{(\binom{τ m + δ}{k})}{(\binom{m + k - 1}{k})},

we can establish, through telescoping, the following formula:

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{τ m + δ}{k})}{(\binom{m + k}{k})} = \frac{m}{τ m + m + δ} + {(- 1)}^{m} \frac{2 m + 1}{τ m + m + δ} \frac{(\binom{τ m + δ}{m + 1})}{(\binom{2 m + 1}{m + 1})} . \end{matrix}

(22)

Analogously, from the relation

\frac{(\binom{m + k}{k})}{(\binom{τ m + δ}{k})} = \frac{τ m + δ + 1}{τ m + m + δ + 2} (J_{k + 1} + J_{k}) - \frac{m + 1}{τ m + m + δ + 2} (J_{k + 1} + J_{k}),

where

J_{k} = \frac{(\binom{m + k}{k})}{(\binom{τ m + δ}{k})} and J_{k} = \frac{(\binom{m + k}{k - 1})}{(\binom{τ m + δ}{k})} .

we then obtain, by means of telescoping, the formula below:

\begin{matrix} \sum_{k = 0}^{m} {(- 1)}^{k} \frac{(\binom{m + k}{k})}{(\binom{τ m + δ}{k})} = \frac{τ m + δ + 1}{τ m + m + δ + 2} + {(- 1)}^{m} \frac{τ m - m + δ}{τ m + m + δ + 2} \frac{(\binom{2 m + 1}{m + 1})}{(\binom{τ m + δ}{m + 1})} . \end{matrix}

(23)

In this paper, we obtained series of summation formulae concerning binomial quotients, which are difficult to calculate or cannot be calculated using classical Saalschütz, Dixon, Watson and Whipple theorems, as well as other

{}_{3}F_{2}

hypergeometric series [23,25,26,27].

Author Contributions

Conceptualization, D.G.; methodology, D.G. and Y.C.; writing—original draft preparation, Y.C.; writing—review and editing, D.G. and Y.C.; supervision, D.G All authors have read and agreed to the published version of the manuscript.

Funding

This work was supported by the Zhoukou Normal University high-level talents start-up fund research project (China, ZKNUC2022007) and the Postgraduate Research & Practice Innovation Program of Jiangsu Province (KYCX24_0725).

Data Availability Statement

All data generated or analyzed during this study are included in this manuscript.

Conflicts of Interest

The authors declare that they have no conflicts interest regarding the publication of this paper.

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Chen, Y.; Guo, D. Combinatorial Identities Concerning Binomial Quotients. Symmetry 2024, 16, 746. https://doi.org/10.3390/sym16060746

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Chen, Yulei, and Dongwei Guo. 2024. "Combinatorial Identities Concerning Binomial Quotients" Symmetry 16, no. 6: 746. https://doi.org/10.3390/sym16060746

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Combinatorial Identities Concerning Binomial Quotients

Abstract

1. Introduction and Motivation

2. Summation Formulae $Ω_{m} (τ, δ, ε)$

3. Summation Formulae $Φ_{m} (τ, δ, ε)$

4. Summation Formulae $Ψ_{m} (τ, δ, ε)$

5. Double Summation Formulae $P_{λ, ε} (m)$

6. Concluding Comments

Author Contributions

Funding

Data Availability Statement

Conflicts of Interest

References

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Combinatorial Identities Concerning Binomial Quotients

Abstract

1. Introduction and Motivation

2. Summation Formulae Ω m ( τ , δ , ε )

3. Summation Formulae Φ m ( τ , δ , ε )

4. Summation Formulae Ψ m ( τ , δ , ε )

5. Double Summation Formulae P λ , ε ( m )

6. Concluding Comments

Author Contributions

Funding

Data Availability Statement

Conflicts of Interest

References

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2. Summation Formulae $Ω_{m} (τ, δ, ε)$

3. Summation Formulae $Φ_{m} (τ, δ, ε)$

4. Summation Formulae $Ψ_{m} (τ, δ, ε)$

5. Double Summation Formulae $P_{λ, ε} (m)$