Hindawi Publishing Corporation Discrete Dynamics in Nature and Society Volume 2012, Article ID 269847, 15 pages doi:10.1155/2012/269847 Research Article Integral Formulae of Bernoulli Polynomials Dae San Kim, 1 Dmitry V. Dolgy, 2 Hyun-Mee Kim, 3 Sang-Hun Lee, 3 and Taekyun Kim 4 1 Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea 2 Hanrimwon, Kwangwoon University, Seoul 139-701, Republic of Korea 3 Division of General Education, Kwangwoon University, Seoul 139-701, Republic of Korea 4 Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea Correspondence should be addressed to Taekyun Kim, [email protected]Received 24 February 2012; Accepted 10 May 2012 Academic Editor: Lee Chae Jang Copyright q 2012 Dae San Kim et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Recently, some interesting and new identities are introduced in Hwang et al., Communicated. From these identities, we derive some new and interesting integral formulae for the Bernoulli polynomials. 1. Introduction As is well known, the Bernoulli polynomials are defined by generating functions as follows: t e t − 1 e xt ∞ n0 B n xt n n! , 1.1see 1–11. In the special case, x 0,B n 0B n are called the nth Bernoulli numbers. The Euler polynomials are also defined by 2 e t 1 e xt e Ext ∞ n0 E n xt n n! 1.2with the usual convention about replacing E n xby E n xsee 1–11. From 1.1and 1.2,
Transcript
Hindawi Publishing CorporationDiscrete Dynamics in Nature and SocietyVolume 2012, Article ID 269847, 15 pagesdoi:10.1155/2012/269847
Research ArticleIntegral Formulae of Bernoulli Polynomials
Dae San Kim,1 Dmitry V. Dolgy,2 Hyun-Mee Kim,3Sang-Hun Lee,3 and Taekyun Kim4
1 Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea2 Hanrimwon, Kwangwoon University, Seoul 139-701, Republic of Korea3 Division of General Education, Kwangwoon University, Seoul 139-701, Republic of Korea4 Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
Correspondence should be addressed to Taekyun Kim, [email protected]
Received 24 February 2012; Accepted 10 May 2012
Academic Editor: Lee Chae Jang
Copyright q 2012 Dae San Kim et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.
Recently, some interesting and new identities are introduced in (Hwang et al., Communicated).From these identities, we derive some new and interesting integral formulae for the Bernoullipolynomials.
1. Introduction
As is well known, the Bernoulli polynomials are defined by generating functions as follows:
t
et − 1ext =
∞∑
n=0
Bn(x)tn
n!, (1.1)
(see [1–11]). In the special case, x = 0, Bn(0) = Bn are called the nth Bernoulli numbers. TheEuler polynomials are also defined by
2et + 1
ext = eE(x)t =∞∑
n=0
En(x)tn
n!(1.2)
with the usual convention about replacing En(x) by En(x) (see [1–11]). From (1.1) and (1.2),
2 Discrete Dynamics in Nature and Society
we can easily derive the following equation:
t
et − 1ext =
t
2
(2ext
et + 1
)+(
t
et − 1
)(2ext
et + 1
)
=∞∑
n=0
⎛⎜⎝
n∑
k=0k /= 1
(nk
)BkEn−k(x)
⎞⎟⎠
tn
n!.
(1.3)
By (1.1) and (1.3), we get
Bn(x) =n∑
k=0k /= 1
(nk
)BkEn−k(x), (n ∈ Z+ = N ∪ {0}). (1.4)
From (1.1), we have
Bn(x) =n∑
l=0
(nl
)Blx
n−l. (1.5)
Thus, by (1.5), we get
d
dxBn(x) = n
n−1∑
l=0
(n − 1l
)Blx
n−1−l = nBn−1(x). (1.6)
It is known that En(0) = En are called the nth Euler numbers (see [7]). The Euler polynomialsare also given by
En(x) = (E + x)n =n∑
l=0
(nl
)Elx
n−l, (1.7)
(see [6]). From (1.7), we can derive the following equation:
d
dxEn(x) = n
n−1∑
l=0
(nl
)Elx
n−1−l = nEn−1(x). (1.8)
By the definition of Bernoulli and Euler numbers, we get the following recurrence formulae:
where δn,k is the kronecker symbol (see [5]). From (1.6), (1.8), and (1.9), we note that
∫1
0Bn(x)dx =
δ0,nn + 1
,
∫1
0En(x)dx = −2En+1
n + 1, (1.10)
Discrete Dynamics in Nature and Society 3
where n ∈ Z+. The following identity is known in [5]:
m∑
j=0
k∑
l=0
(a + b + 1)m−j(a + 1)k−l(mj
)(kl
)(−1)j+lj + l + 1
+m∑
j=0
k∑
l=0
((a + b + 1)m−j(a + 1)k−l − (a + b)m−jak−l
)(mj
)(kl
)Bj+l+1(x)j + l + 1
= (x + a)k(x + a + b)m, where a, b ∈ Z.
(1.11)
From the identities of Bernoulli polynomials, we derive some new and interesting integralformulae of an arithmetical nature on the Bernoulli polynomials.
2. Integral Formulae of Bernoulli Polynomials
From (1.1) and (1.2), we note that
2et + 1
ext =1t
(2(et − 1
)
et + 1
)(text
et − 1
)
=1t
(2 − 2
2et + 1
)(text
et − 1
)
=1t
(−
∞∑
l=1
2El
l!tl)( ∞∑
m=0
Bm(x)tm
m!
)
= −2( ∞∑
l=0
El+1
l + 1tl
l!
)( ∞∑
m=0
Bm(x)tm
m!
)
= −2∞∑
n=0
(n∑
l=0
El+1
l + 1
(nl
)Bn−l(x)
)tn
n!.
(2.1)
Therefore, by (1.2) and (2.1), we obtain the following theorem.
Theorem 2.1. For n ∈ Z+, one has
En(x) = −2n∑
l=0
(nl
)El+1
l + 1Bn−l(x). (2.2)
Let us take the definite integral from 0 to 1 on both sides of (1.4): for n ≥ 2,
0 = −2n∑
k=0k /= 1
(nk
)Bk
En−k+1n − k + 1
= −2BnE1 − 2n−1∑
k=0k /= 1
(nk
)Bk
En−k+1n − k + 1
. (2.3)
4 Discrete Dynamics in Nature and Society
By (2.3), we get
Bn = 2n−1∑
k=0k /= 1
(nk
)BkEn−k+1n − k + 1
. (2.4)
Therefore, by (2.4), we obtain the following theorem.
Theorem 2.2. For n ∈ N, with n ≥ 2, one has
Bn = 2n−1∑
k=0k /= 1
(nk
)BkEn−k+1n − k + 1
. (2.5)
Let us take k = m, a = 0, and b = −2 in (1.11). Then we have
m∑
j=0
m∑
l=0
(−1)m−j(mj
)(ml
)(−1)j+lj + l + 1
+m∑
j=0
m∑
l=0
(−1)m−j(mj
)(ml
)Bj+l+1(x)j + l + 1
−m∑
j=0(−2)m−j
(mj
)Bj+m+1(x)j +m + 1
= xm(x − 2)m.
(2.6)
It is easy to show that
∫1
0xm(x − 2)mdx = 2
∫1/2
0(2t − 2)m(2t)mdt
= (−1)m22m(2∫1/2
0tm(1 − t)mdt
)= (−1)m22m
∫1
0tm(1 − t)mdt
= (−1)m22m m!m!(2m + 1)!
=(−1)m22m2m + 1
1(2mm
) .
(2.7)
Let us consider the integral from 0 to 1 in (2.6):
m∑
j=0
m∑
l=0
(−1)m−l(mj
)(ml
)1
j + l + 1=
(−1)m22m(2m + 1)
(2mm
) (m ∈ N). (2.8)
Discrete Dynamics in Nature and Society 5
By (2.6) and (2.8), we get
m∑
j=0
m∑
l=0
(−1)m−j(mj
)(ml
)Bj+l+1
j + l + 1
= 2m∑
j=0(−2)m−j
(mj
)1
j +m + 1
j+m∑
k=0k /= 1
(j +m + 1
k
)BkEj+m+2−kj +m + 2 − k
+(−1)m+122m
2m + 11(2mm
) , for m ∈ N.
(2.9)
Therefore, by (2.9), we obtain the following theorem.
Theorem 2.3. For m ∈ N, one has
m∑
j=0
m∑
l=0
(−1)m−j(mj
)(ml
)Bj+l+1
j + l + 1
= 2m∑
j=0(−2)m−j
(mj
)1
j +m + 1
j+m∑
k=0k /= 1
(j +m + 1
k
)BkEj+m+2−kj +m + 2 − k
+(−1)m+122m
2m + 11(2mm
) .
(2.10)
Lemma 2.4. Let a, b ∈ Z. For m, k ∈ Z+, one has
m∑
j=0
k∑
l=0
(a + b + 1)m−j(a + 1)k−l(mj
)(kl
)Ej+l(x)
+m∑
j=0
k∑
l=0
(a + b)m−jak−l(mj
)(kl
)Ej+l(x) = 2(x + a + b)m(x + a)k,
(2.11)
(see [5]).
Let us take k = m, a = 1, b = −2 in Lemma 2.4. Then we have
m∑
l=0
2m−l(ml
)Em+l(x) +
m∑
j=0
m∑
l=0
(−1)m−j(mj
)(ml
)Ej+l(x) = 2
(x2 − 1
)m. (2.12)
Taking integral from 0 to 1 in (2.12), we get
−2m∑
l=0
2m−l(ml
)Em+l+1
m + l + 1− 2
m∑
j=0
m∑
l=0
(−1)m−l(mj
)(ml
)Ej+l+1
j + l + 1= 2∫1
0
(x2 − 1
)mdx. (2.13)
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It is easy to show that
∫1
0
(x2 − 1
)mdx = (−1)m
m∏
k=1
(2k
2k + 1
)=
(−1)m22m(2m + 1)
(2mm
) . (2.14)
Thus, by (2.13) and (2.14), we get
m∑
j=0
m∑
l=0
(−1)m−l(mj
)(ml
)Ej+l+1
j + l + 1= −
m∑
l=0
2m−l(ml
)Em+l+1
m + l + 1+
(−1)m+122m
(2m + 1)(2mm
) . (2.15)
Therefore, by (2.2) and (2.15), we obtain the following theorem.
Theorem 2.5. For m ∈ Z+, one has
m∑
j=0
m∑
l=0
(−1)m−l(mj
)(ml
)Ej+l+1
j + l + 1+
(−1)m22m(2m + 1)
(2mm
)
= 2m∑
l=0
2m−l(ml
)1
m + l + 1
m+l+1∑
k=0
(m + l + 1
k
)Ek+1
k + 1Bm+l+1−k.
(2.16)
3. p-Adic Integral on Zp Associated with Bernoulli and Euler Numbers
Let p be a fixed odd prime number. Throughout this section, Zp, Qp, and Cp will denote thering of p-adic integers, the field of p-adic rational numbers, and the completion of algebraicclosure of Qp, respectively. Let νp be the normalized exponential valuation of Cp with |p|p =p−νp(p) = 1/p. Let UD(Zp) be the space of uniformly differentiable functions on Zp. For f ∈UD(Zp), the bosonic p-adic integral on Zp is defined by
I(f)=∫
Zp
f(x)dμ(x) = limn→∞
1pn
pn−1∑
x=0
f(x), (3.1)
(see [8]). Thus, by (3.1), we get
∫
Zp
f1(x)dμ(x) =∫
Zp
f(x)dμ(x) + f ′(0), (3.2)
where f1(x) = f(x + 1), and f ′(0) = df(x)/dx|x=0. Let us take f(y) = et(x+y). Then we have
∫
Zp
et(x+y)dμ(y)=
t
et − 1etx =
∞∑
n=0
Bn(x)tn
n!. (3.3)
Discrete Dynamics in Nature and Society 7
From (3.3), we have
∫
Zp
(x + y
)ndμ(y)= Bn(x),
∫
Zp
yndμ(y)= Bn. (3.4)
From (1.2), we can derive the following integral equation:
I(fn)= I(f)+
n−1∑
i=0
f ′(i) (n ∈ N). (3.5)
Thus, from (3.4) and (3.5), we get
∫
Zp
(x + n)mdμ(x) =∫
Zp
xmdμ(x) +mn−1∑
i=0
im−1. (3.6)
From (3.6), we have
Bm(n) − Bm = mn−1∑
i=0
im−1 (m ∈ Z+). (3.7)
The fermionic p-adic integral on Zp is defined by Kim as follows [6, 7]:
I−1(f)=∫
Zp
f(x)dμ−1(x) = limn→∞
pn−1∑
x=0
f(x)(−1)x. (3.8)
Let f1(x) = f(x + 1). Then we have
I−1(f1)= − I−1
(f)+ 2f(0),
I−1(f2)= − I−1
(f1)+ 2f1(0) = −I−1
(f1)+ 2f(1)
= (−1)2I−1(f)+ 2(−1)2−1f(0) + 2f(1).
(3.9)
Continuing this process, we obtain the following equation:
I−1(fn)= (−1)nI−1
(f)+ 2
n−1∑
l=0
(−1)n−l−1f(l), where fn(x) = f(x + n). (3.10)
Thus, by (3.10), we have
∫
Zp
(x + n)mdμ−1(x) = (−1)n∫
Zp
xmdμ−1(x) + 2n−1∑
l=0
(−1)n−l−1lm. (3.11)
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Let us take f(y) = et(x+y). By (3.9), we get
∫
Zp
et(x+y)dμ−1(y)=
2ext
et + 1=
∞∑
n=0
En(x)tn
n!. (3.12)
From (3.2), we have theWitt’s formula for the nth Euler polynomials and numbers as follows:
∫
Zp
(x + y
)ndμ−1
(y)= En(x),
∫
Zp
yndμ−1(y)= En, where n ∈ Z+. (3.13)
By (3.11) and (3.13), we get
Em(n) = (−1)n(Em + 2
n−1∑
l=0
(−1)l−1lm), (m ∈ Z+, n ∈ N). (3.14)
Let us consider the following p-adic integral on Zp:
K1 =∫
Zp
Bn(x)dμ(x) =n∑
l=0
(nl
)Bn−l
∫
Zp
xldμ(x)
=n∑
l=0
(nl
)Bn−lBl.
(3.15)
From (1.4) and (3.15), we have
K1 =n∑
k=0k /= 1
(nk
)Bk
n−k∑
l=0
En−k−l
(n − kl
)∫
Zp
xldμ(x)
=n∑
k=0k /= 1
n−k∑
l=0
(nk
)(n − kl
)BkBlEn−k−l.
(3.16)
Therefore, by (3.15) and (3.16), we obtain the following theorem.
Theorem 3.1. For n ∈ Z+, one has
n∑
l=0
(nl
)Bn−lBl =
n∑
k=0k /= 1
n−k∑
l=0
(nk
)(n − kl
)BkBlEn−k−l. (3.17)
Now, we set
K2 =∫
Zp
Bn(x)dμ−1(x) =n∑
l=0
(nl
)Bn−lEl. (3.18)
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By (1.4), we get
K2 =n∑
k=0k /= 1
(nk
)Bk
n−k∑
l=0
En−k−l
(n − kl
)∫
Zp
xldμ−1(x)
=n∑
k=0k /= 1
n−k∑
l=0
(nk
)(n − kl
)BkEn−k−lEl.
(3.19)
Therefore, by (3.18) and (3.19), we obtain the following theorem.
Theorem 3.2. For n ∈ Z+, one has
n∑
l=0
(nl
)Bn−lEl =
n∑
k=0k /= 1
n−k∑
l=0
(nk
)(n − kl
)BkEn−k−lEl. (3.20)
Let us consider the following integral on Zp:
K3 =∫
Zp
En(x)dμ−1(x) =n∑
l=0
(nl
)En−l
∫
Zp
xldμ−1(x) =n∑
l=0
(nl
)En−lEl. (3.21)
From (2.2), we have
K3 = −2n∑
l=0
El+1
l + 1
(nl
)n−l∑
k=0
(n − lk
)Bn−l−k
∫
Zp
xkdμ−1(x)
= −2n∑
l=0
n−l∑
k=0
(nl
)(n − lk
)El+1
l + 1Bn−l−kEk.
(3.22)
Therefore, by (3.21) and (3.22), we obtain the following theorem.
Theorem 3.3. For n ∈ Z+, one has
n∑
l=0
(nl
)En−lEl = −2
n∑
l=0
n−l∑
k=0
(nl
)(n − lk
)El+1
l + 1EkBn−l−k. (3.23)
Now, we set
K4 =∫
Zp
En(x)dμ(x) =n∑
l=0
(nl
)En−lBl. (3.24)
10 Discrete Dynamics in Nature and Society
By (2.2), we get
K4 = −2n∑
l=0
n−l∑
k=0
(nl
)(n − lk
)El+1
l + 1Bn−l−kBk. (3.25)
Therefore, by (3.24) and (3.25), we obtain the following corollary.
Corollary 3.4. For n ∈ Z+, we have
n∑
l=0
(nl
)ElBl = −2
n∑
l=0
n−l∑
k=0
(nl
)(n − lk
)El+1
l + 1Bn−l−kBk. (3.26)
Let us assume that a, b, c, d ∈ Z. From Lemma 2.4 and (3.13), we note that
∫
Zp
((a + b + 1) +
(x + y
))m((a + 1) +(x + y
))kdμ−1
(y)
+∫
Zp
((a + b) +
(x + y
))m((a + (x + y))kdμ−1
(y)
= 2(x + a + b)m(x + a)k.
(3.27)
By (3.27), we get
2(x + a + b)m(x + a)k =∫
Zp
((a + b − c + 1) +
(x + c + y
))m((a − c + 1) +(x + c + y
))kdμ−1
(y)
+∫
Zp
((a + b − d) +
(x + y + d
))m((a − d) +(x + y + d
))kdμ−1
(y).
(3.28)
Thus, by (3.28) and (3.13), we obtain the following lemma (see [5]).
Lemma 3.5. Let a, b, c, d ∈ Z. For m, k ∈ Z+, one has
m∑
j=0
k∑
l=0
(mj
)(kl
)(a + b − c + 1)m−j(a − c + 1)k−lEj+l(x + c)
+m∑
j=0
k∑
l=0
(a + b − d)m−j(a − d)k−l(mj
)(kl
)Ej+l(x + d) = 2(x + a + b)m(x + a)k.
(3.29)
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Let us consider the formula in Lemma 3.5 with d = c − 1. Then we have
m∑
j=0
k∑
l=0
(a + b − c + 1)m−j(a − c + 1)k−l(mj
)(kl
)(Ej+l(x + c) + Ej+l(x + c − 1)
)
= 2(x + a + b)m(x + a)k.
(3.30)
Taking∫Zp
dμ(x) on both sides of (3.30),
LHS =m∑
j=0
k∑
l=0
(a + b − c + 1)m−j(a − c + 1)k−l(mj
)(kl
) j+l∑
s=0
(j + ls
)
× Ej+l−s
∫
Zp
((x + c)s + (x + c − 1)s
)dμ(x)
= 2m∑
j=0
k∑
l=0
(a + b − c + 1)m−j(a − c + 1)k−l(mj
)(kl
) j+l∑
s=0
(j + ls
)
× Ej+l−sBs(c − 1) +m∑
j=0
k∑
l=0
(a + b − c + 1)m−j(a − c + 1)k−l(mj
)(kl
)
× (j + l)Ej+l−1(c − 1).
(3.31)
By the same method, we get
RHS = 2m∑
s=0
(ms
)bm−s
∫
Zp
(x + a)s+kdμ(x)
= 2m∑
s=0
(ms
)bm−sBs+k(a).
(3.32)
Therefore, by (3.31) and (3.32), we obtain the following proposition.
Proposition 3.6. Let a, b, c ∈ Z. Then one has
2m∑
j=0
k∑
l=0
(a + b − c + 1)m−j(a − c + 1)k−l(mj
)(kl
) j+l∑
s=0
(j + ls
)Ej+l−s
× Bs(c − 1) +m∑
j=0
k∑
l=0
(a + b − c + 1)m−j(a − c + 1)k−l(mj
)(kl
)(j + l)Ej+l−1(c − 1)
= 2m∑
s=0
(ms
)bm−sBs+k(a).
(3.33)
12 Discrete Dynamics in Nature and Society
Replacing c by c + 1, we have
2m∑
j=0
k∑
l=0
(a + b − c)m−j(a − c)k−l(mj
)(kl
) j+l∑
s=0
(j + ls
)Ej+l−sBs(c)
+m∑
j=0
k∑
l=0
(j + l)(a + b − c)m−j(a − c)k−l
(mj
)(kl
)Ej+l−s(c)
= 2m∑
s=0
(ms
)bm−sBs+k(a).
(3.34)
From (3.4) and (3.7), we derive some identity for the first term of the LHS of (3.34).The first term of the LHS of (3.34)
= 2m∑
j=0
k∑
l=0
(a + b − c)m−j(a − c)k−l(mj
)(kl
) j+l∑
s=0
(j + ls
)Ej+l−s
×(Bs + s
c−1∑
i=0
is−1)
= 2m∑
j=0
k∑
l=0
(a + b − c)m−j(a − c)k−l(mj
)(kl
) j+l∑
s=0
(j + ls
)Ej+l−sBs
+ 2c−1∑
i=0
m∑
j=0
k∑
l=0
(a + b − c)m−j(a − c)k−l(mj
)(kl
)(j + l)(−1)0
×(Ej+l−1 + 2
i−1∑
e=0(−1)e−1ej+l−1
)
= 2m∑
j=0
k∑
l=0
j+l∑
s=0
(mj
)(kl
)(j + ls
)(a + b − c)m−j(a − c)k−l
× Ej+l−sBs + 2mm−1∑
j=0
k∑
l=0
(m − 1j
)(kl
)(a + b − c)m−1−j(a − c)k−lEj+lδc≡1
+ 2km∑
j=0
k−1∑
l=0
(mj
)(k − 1l
)(a + b − c)m−j(a − c)k−1−lEj+lδc≡1
+ 4mc−2∑
e=0(a + b − c + e)m−1(a − c + e)kδc≡e
+ 4kc−2∑
e=0(a + b − c + e)m(a − c + e)k−1δc≡e,
(3.35)
Discrete Dynamics in Nature and Society 13
where
δc≡k =
{1 if c ≡ k(mod 2),0 if c /= k(mod 2).
(3.36)
The second term of the LHS of (3.34)
=m∑
j=0
k∑
l=0
(j + l)(a + b − c)m−j(a − c)k−l
(mj
)(kl
)(−1)c
(Ej+l−1 + 2
c−1∑
i=0(−1)i−1ij+l−1
)
= (−1)cm∑
j=0
k∑
l=0
(j + l)(a + b − c)m−j(a − c)k−l
(mj
)(kl
)Ej+l−1
+ 2(−1)cc−1∑
i=0(−1)i−1m(a + b − c + i)m−1(a − c + i)k
+ 2(−1)cc−1∑
i=0(−1)i−1(a + b − c + i)mk(a − c + i)k−1
= (−1)cm∑
j=0
k∑
l=0
(j + l)(a + b − c)m−j(a − c)k−l
(mj
)(kl
)Ej+l−1
+ 2m(−1)cc−1∑
i=0(−1)i−1(a + b − c + i)m−1(a − c + i)k
+ 2k(−1)cc−1∑
i=0(−1)i−1(a + b − c + i)m(a − c + i)k−1.
(3.37)
Therefore, by (3.34), (3.35), and (3.37), we obtain the following theorem.
Theorem 3.7. Let a, b, c ∈ Z with c ≥ 1. Then one has
2m∑
j=0
k∑
l=0
j+l∑
s=0
(mj
)(kl
)(j + ls
)(a + b − c)m−j(a − c)k−lEj+l−sBs
+ 2mδc≡1m∑
j=0
k∑
l=0
(m − 1j
)(kl
)(a + b − c)m−1−j(a − c)k−lEj+l
+ 2kδc≡1m∑
j=0
k∑
l=0
(mj
)(k − 1l
)(a + b − c)m−j(a − c)k−1−lEj+l
+ (−1)cm∑
j=0
k∑
l=0
(j + l)(a + b − c)m−j(a − c)k−l
(mj
)(kl
)Ej+l−1
14 Discrete Dynamics in Nature and Society
+ 2mc−1∑
e=0(a + b − c + e)m−1(a − c + e)k
+ 2kc−1∑
e=0(a + b − c + e)m(a − c + e)k−1
= 2m∑
s=0
(ms
)bm−sBs+k(a),
(3.38)
where
δc≡k =
{1 if c ≡ k(mod 2),0 if c /≡ k(mod 2).
(3.39)
Remark 3.8. Here, we note that
4mc−2∑
e=0(a + b − c + e)m−1(a − c + e)kδc≡e + 2m(−1)c
c−1∑
i=0(−1)i−1(a + b − c + i)m−1(a − c + i)k
= 2mc−1∑
e=0(a + b − c + e)m−1(a − c + e)k.
(3.40)
Acknowledgment
The first author was supported by National Research Foundation of Korea Grant funded bythe Korean Government 2011-0002486.
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