1 Introduction

Let \(\mathbb{N}\) be the set of nonnegative integers and \(\mathbb {N}_{+}=\mathbb{N}\setminus\{0\}\). Throughout this note, we assume that \(m,n\in\mathbb{N}\), \(x\in\mathbb{R}\), and that \(f:\mathbb {R}\to\mathbb{R}\) is an arbitrary function. The kth forward differences of f are recursively defined by \(\Delta^{0}f(x)=f(x)\), \(\Delta^{1} f(x)=f(x+1)-f(x)\), and

$$ \Delta^{k} f(x)=\Delta^{1} \bigl( \Delta^{k-1} f \bigr) (x)=\sum_{i=0}^{k} \binom{k}{i}(-1)^{k-i} f(x+i),\quad k\in\mathbb{N}_{+}. $$
(1)

The starting point of this note is the following classical formula for sums of powers:

$$ \sum_{k=0}^{n} (x+k)^{m}=\frac{B_{m+1}(x+n+1)-B_{m+1}(x)}{m+1}, $$
(2)

where \(B_{m}(x)\) is the mth Bernoulli polynomial. Since the time of James Bernoulli (1655-1705), several methods have been developed to find such sums, trying in many occasions to obtain different generalizations. For instance, Kannappan and Zhang [1] (see also the references therein) have used Cauchy’s equation to prove (2), when the monomial function \(x^{m}\) is replaced by a polynomial of degree m. Some q-analogues of formula (2) can be found in Guo and Zeng [2] and the references therein.

On the other hand, the sums in (2) can also be computed by means of the forward differences of the monomial function \(\psi _{m}(u)=u^{m}\), \(u\in\mathbb{R}\). We actually have (see, for instance, Rosen [3], p.199, or Spivey [4])

$$ \sum_{k=0}^{n} (x+k)^{m}=\sum_{k=0}^{m\wedge n} \binom {n+1}{k+1}\Delta^{k} \psi_{m}(x), $$
(3)

where \(m\wedge n=\min(m,n)\). For \(x=0\), formula (3) can be written in terms of the Stirling numbers of the second kind \(S(m,k)\) defined as

$$ S(m,k)= \frac{1}{k!} \Delta^{k} \psi_{m}(0). $$

Computationally, formulas (2) and (3) are equivalent in the sense that the computation of a sum of \(n+1\) terms is reduced to the computation of a polynomial in n of degree \(m+1\). However, (2) can easily be derived from (3) as follows. Suppose that \((P_{m}(x))_{m\geq0}\) is a sequence of polynomials satisfying

$$ \Delta^{1} P_{m+1}(x)=c_{m} \psi_{m}(x) $$
(4)

for a certain constant \(c_{m}\) only depending upon m. Then we have from (3), (4), and formula (7) below

$$\begin{aligned} \sum_{k=0}^{n} (x+k)^{n}&= \sum_{k=0}^{n} \binom{n+1}{k+1}\Delta ^{k} \psi_{m}(x) \\ &={\frac{1}{c_{m}}} \sum_{k=0}^{n} \binom{n+1}{k+1} \Delta^{k+1} P_{m+1}(x) \\ &=\frac{P_{m+1}(x+n+1)-P_{m+1}(x)}{c_{m}}. \end{aligned}$$

The Bernoulli polynomials satisfy (4) with \(c_{m}={m+1}\). However, one can construct other sequences of polynomials \((P_{m}(x))_{m\geq0}\) fulfilling (4) (in this respect, see Luo et al. [5]). For this reason, we will extend formula (3) rather than (2). This is done in Theorem 2.1 below by means of a simple identity involving binomial mixtures.

2 Main results

Let \(\mathbb{S}_{n}=(S_{n}(t), 0\leq t \leq1)\) be a stochastic process such that \(S_{n}(t)\) has the binomial law with parameters n and t, i.e.,

$$ P \bigl(S_{n}(t)=k \bigr)=\binom{n}{k}t^{k} (1-t)^{n-k},\quad k=0,1,\ldots,n, $$
(5)

and let T be a random variable taking values in \([0,1]\) and independent of \(\mathbb{S}_{n}\). The random variable \(S_{n}(T)\), obtained by randomizing the success parameter t by T, is called a binomial mixture with mixing random variable T (see [6] and the references therein). As follows from (5), the probability law of \(S_{n}(T)\) is given by

$$ P \bigl(S_{n}(T)=k \bigr)=\binom{n}{k}E { \bigl[{T^{k} (1-T)^{n-k}} \bigr]},\quad k=0,1,\ldots,n, $$

where E stands for mathematical expectation. Our first main result is the following.

Theorem 2.1

With the preceding notations, we have

$$ E{ \bigl[{f \bigl(x+S_{n}(T) \bigr)} \bigr]}=\sum _{k=0}^{n} \binom{n}{k} f(x+k) E{ \bigl[{T^{k} (1-T)^{n-k}} \bigr]} =\sum _{k=0}^{n} \binom{n}{k}\Delta^{k} f(x)E{ \bigl[{T^{k}} \bigr]}. $$

Let U be a random variable having the uniform distribution on \((0,1)\). Observe that

$$ E{ \bigl[{U^{k}(1-U)^{n-k}} \bigr]}= \int_{0}^{1} \theta^{k} (1-\theta )^{n-k}\, d\theta= \beta(k+1,n-k+1)=\frac{k! (n-k)!}{(n+1)!}, $$
(6)

where \(\beta(\cdot, \cdot)\) is Euler’s beta function. Setting \(T=U\) and \(f=\psi_{m}\) in Theorem 2.1, we obtain (3), as follows from (6) and the fact that \(\Delta^{k} \psi_{m} (x)=0\), \(k=m+1,m+2,\ldots\) . On the other hand, choosing \(T=1\) in Theorem 2.1, we obtain the well-known identity (see, for instance, Flajolet and Vepstas [7])

$$ f(x+n)=\sum_{k=0}^{n} \binom{n}{k}\Delta^{k} f(x). $$
(7)

In the terminology of binomial transforms (see, for instance, Mu [8] and the references therein), identity (7) means that \((f(x+k))_{k\geq0}\) is the binomial transform of \((\Delta^{k} f(x))_{k\geq0}\). In this sense, Theorem 2.1 appears as a generalization of (7).

Every choice of the function f and the random variable T in Theorem 2.1 gives us a different binomial identity. Whenever the probability density of T includes the uniform density on \((0,1)\) as a particular case, we are able to obtain a different extension of formula (3). In this respect, we give the following two corollaries of Theorem 2.1.

Corollary 2.2

For any \(p>0\) and \(q>0\), we have

$$ \sum_{k=0}^{n} \binom{-p}{k} \binom{-q}{n-k}f(x+k)=\sum_{k=0}^{n} \binom{-p}{k}\binom{-(p+q+k)}{n-k}\Delta^{k} f(x). $$

Finally, recall that the discrete Cesàro operator C is defined as

$$ Cf(x+n)=\frac{f(x)+f(x+1)+\cdots+f(x+n)}{n+1}. $$
(8)

We denote by \(C^{j}\) the j iterate of C, \(j\in\mathbb{N}_{+}\) (see Galaz and Solís [9] and Adell and Lekuona [10] for the asymptotic behavior of such iterates, as \(j\to \infty\)).

Corollary 2.3

For any \(j\in\mathbb{N}_{+}\), we have

$$\begin{aligned} C^{j}f(x+n)&=\sum_{k=0}^{n} f(x+k)\binom{n}{k} \int_{0}^{1} \theta ^{k} (1- \theta)^{n-k} \frac{(-\log\theta)^{j-1}}{(j-1)!}\, d\theta \\ &=\sum_{k=0}^{n} \binom{n}{k} \frac{1}{(k+1)^{j}}\Delta^{k} f(x). \end{aligned}$$

Observe that both corollaries extend formula (3) by choosing \(f=\psi_{m}\) and \(p=q=1\) in Corollary 2.2, and \(f=\psi_{m}\) and \(j=1\) in Corollary 2.3.

3 The proofs

Proof of Theorem 2.1

Let \(t\in[0,1]\). We have from (1) and (5)

$$\begin{aligned} E{ \bigl[{f \bigl(x+S_{n}(t) \bigr)} \bigr]}&= \sum _{k=0}^{n} \binom{n}{k} f(x+k) t^{k} (1-t)^{n-k} \\ &= \sum_{k=0}^{n} \binom{n}{k} f(x+k) t^{k} \sum_{j=0}^{n-k} \binom {n-k}{j}(-t)^{j} \\ &{ =\sum_{s=0}^{n} \sum _{k+j=s} \binom{n}{k} \binom{n-k}{j}(-1)^{j} f(x+k)t^{k+j}} \\ &=\sum_{s=0}^{n} \binom{n}{s} t^{s} {\sum_{k=0}^{s}} \binom{s}{k} {(-1)^{s-k}f(x+k)} \\ &=\sum_{s=0}^{n} \binom{n}{s} t^{s} \Delta^{s} f(x). \end{aligned}$$

Thus, it suffices to replace t by the random variable T and then to take expectations. □

Proof of Corollary 2.2

Let T be a random variable having the beta density

$$ \rho(\theta)=\frac{\theta^{p-1}(1-\theta)^{q-1}}{\beta (p,q)},\quad\theta\in(0,1),p>0, q>0. $$

As in (6), we have

$$ E{ \bigl[{T^{r}(1-T)^{s}} \bigr]}= \frac{1}{\beta(p,q)} \int_{0}^{1} \theta^{p+r-1}(1- \theta)^{q+s-1}\, d\theta=\frac{\beta (p+r,q+s)}{\beta(p,q)}, $$
(9)

whenever \(r>-p\) and \(s>-q\). Hence, applying Theorem 2.1, we get

$$ \sum_{k=0}^{n} \binom{n}{k} \beta(p+k, q+n-k) f(x+k)= \sum_{k=0}^{n} \binom{n}{k}\beta(p+k,q) \Delta^{k} f(x). $$
(10)

The conclusion follows from (10) and the well-known formulas

$$ \beta(p,q)=\frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)},\qquad\frac {\Gamma(p+k)}{k! \Gamma(p)}=(-1)^{k} \binom{-p}{k},\quad k\in\mathbb{N}. $$

 □

Proof of Corollary 2.3

Let \(j\in\mathbb{N}_{+}\). The following formula for the j iterate of the discrete Cesàro operator C was shown by Hardy [11], Section II.12,

$$ C^{j}f(x+n)=\sum_{k=0}^{n} f(x+k) \binom{n}{k} \int_{0}^{1} \theta ^{k} (1- \theta)^{n-k} \frac{(-\log\theta)^{j-1}}{(j-1)!}\, d\theta. $$
(11)

A probabilistic representation of (11) can be built as follows (see [10] for more details). Let \((U_{k})_{k\geq1}\) be a sequence of independent identically distributed random variables having the uniform distribution on \((0,1)\), and denote \(T_{j}=U_{1}\cdots U_{j}\). It turns out (cf. [10], Lemma 2.2) that the probability density of \(T_{j}\) is given by

$$ \rho_{j}(\theta)=\frac{(-\log\theta)^{j-1}}{(j-1)!},\quad0< \theta< 1. $$
(12)

On the other hand, we see that

$$ E{ \bigl[{T_{j}^{k}} \bigr]}=E{ \bigl[{U_{1}^{k}} \bigr]}\cdots E{ \bigl[{U_{j}^{k}} \bigr]}=\frac{1}{(k+1)^{j}},\quad k\in\mathbb{N}. $$
(13)

Therefore, the conclusion follows by choosing \(T=T_{j}\) in Theorem 2.1 and taking into account (11)-(13). □