EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 10, No. 3, 2017, 506-515
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On identities for sequences of binomial sums with the
terms of sequences {ukn} and {vkn}

Neşe Ömür1,∗, Cemile Duygu Şener1

1 Mathematics Department, Science and Art Faculty, Kocaeli University, Turkey

Abstract. In this paper, considering technique used in [4], and the sequences {ukn} and {vkn} ,
we derive the sequences {gkn} and {hkn} . Also with the aid of generating matrix for the terms of
these sequences for a positive integer k, we derive some combinatorial identities for the sequence
{gkn} .
2010 Mathematics Subject Classifications: 11B39, 05A10, 05A15, 05A19

Key Words and Phrases: Binomial sums, generalized Fibonacci numbers, recurrence relation

1. Introduction

Matrix methods are very convenient for deriving certain of linear recurrence sequences.
Some authors have used matrix methods of other methods to derive some identities, com-
binatorial representations of linear recurrence relations etc[3, 6, 10, 13, 14, 15, 16, 8, 9].

In [13], the author gives a new formula for the nth power of an arbitrary 2× 2 matrix
and derive various matrix identities and formulae for the nth power of particular matrices
to obtain various combinatorial identities. The generalized second order sequences {un}
and {vn} , are defined for n > 0 and nonzero integer numbers p, q by

un+1 = pun + qun−1 and vn+1 = pvn + qvn−1

in which u0 = 0, u1 = 1 and v0 = 2, v1 = p, respectively. When p = q = 1, un = Fn (the
nth Fibonacci number) and vn = Ln (the nth Lucas number).

If α and β are the roots of equation x2−px−q = 0, the Binet formulae of the sequences
{un} and {vn} have the form

un =
αn − βn

α− β
and vn = αn + βn,

∗Corresponding author.

Email addresses: neseomur@kocaeli.edu.tr (N. Ömür,) cemileduygusener@gmail.com (C.D.Şener)

http://www.ejpam.com 506 c© 2017 EJPAM All rights reserved.



N. Ömür, C. D. Şener / Eur. J. Pure Appl. Math, 10 (3) (2017), 506-515 507

respectively. From [7], E. Kılıç and P. Stanica derived the following recurrence relations
for the sequences {ukn} and {vkn} for k ≥ 0, n > 0. It is clearly that

uk(n+1) = vkukn + (−1)k+1 qkuk(n−1) and vk(n+1) = vkvkn + (−1)k+1 qkvk(n−1),

where the initial conditions of the sequences {ukn} and {vkn} are 0, uk, and 2, vk, respec-
tively. The Binet formulae of the sequences {ukn} and {vkn} are given by

ukn =
αkn − βkn

α− β
and vkn = αkn + βkn,

respectively. From the Binet formulas, one can see that u−kn = (−1)kn+1 ukn and u2kn =
uknvkn.

In [1] and [2], the authors obtained some new identities for the sequence {un}. For
example, for n ≥ 1,

n∑
k=0

(
n

k

)(
α

q

)k

uk =
α

q

(
pα

q
+ 2

)n−1
,

and
n∑

k=0

(
n

k

)(
β

q

)k

uk =
β

q

(
pβ

q
+ 2

)n−1
.

Let {ak} and {bk} be sequences with the property that ak is the finite difference of bk,
that is, ak = ∆bk := bk+1 − bk, for k ≥ 0. We take

gn =
n∑

k=0

(
n

k

)
ak and hn =

n∑
k=0

(
n

k

)
bk. (1)

In [11], Komatsu obtained several sequences of binomial sums of generalized Fibonacci
numbers. For example,

n∑
k=0

(
n

k

)
ckuk = rn (n ≥ 0)

satisfies the recurrence relation

rn = (ac+ 2) rn−1 +
(
bc2 − ac− 1

)
rn−2 (n ≥ 2)

with r0 = 0, r1 = c and
n∑

k=0

(
n

k

)
cn−kdkuk = λn (n ≥ 0)

satisfies the recurrence relation

λn = (ad+ 2c)λn−1 +
(
bd2 − acd− c2

)
λn−2 (n ≥ 2)

with initial conditions λ0 = 0 and λ1 = d, where c, d are nonzero real numbers.



N. Ömür, C. D. Şener / Eur. J. Pure Appl. Math, 10 (3) (2017), 506-515 508

In [4], the authors obtain some binomial summation identities of sequences {rn} and
{λn}:

2n∑
k=0

(
2n

k

)
(−1)k

(
bc2 − ac− 1

)2n−k
r2k+1 = (ac+ 2)2n r2n+1,

2n∑
k=0

(
2n

k

)
(−1)k

(
bc2 − ac− 1

)2n−k
r2k = (ac+ 2)2n r2n.

2. Some Results

In this section, firstly, we define sequences {gkn} and {hkn} and then derive some new
combinatorial identities for these sequences.

Lemma 1. For n ≥ 0, the sum
n∑

i=0

(
n
i

)
ckiuki = gkn satisfies the recurrence relation

gk(n+2) =
(
ckvk + 2

)
gk(n+1) −

(
c2k(−q)k + ckvk + 1

)
gkn,

where initial conditions g0 = 0, gk = ckuk.

Proof. Considering akn = cknukn and bkn = ck(n+1)uk(n+1) in (1), the proof is com-
pleted as similar to proof technique in [11].

Lemma 2. The generating function U(z) of
n∑

i=0

(
n
i

)
ckiuki = gkn is given by

U(z) =
zkgk

1− (ckvk + 2) zk + (c2k(−q)k + ckvk + 1) z2k
.

Proof. Observed that

U(z) = g0z
0 + gkz

k + g2kz
2k + ...+ gknz

kn + ...

zkU(z) = g0z
k + gkz

2k + g2kz
3k + ...+ gk(n−1)z

kn + ...

z2kU(z) = g0z
2k + gkz

3k + g2kz
4k + ...+ gk(n−2)z

kn + ...

...

From here, we have

U(z)
(

1−
(
ckvk + 2

)
zk −

(
c2k(−q)k + ckvk + 1

)
z2k
)

= zkgk +

∞∑
i=2

(
gki −

(
ckvk + 2

)
gk(i−1) +

(
c2k(−q)k + ckvk + 1

)
gk(i−2)

)
zki.

From the recurrence relation in Lemma 1, we complete the proof for U(z).

Similarly, the proofs of the following lemmas are given as the proofs of Lemmas 1 and
2.



N. Ömür, C. D. Şener / Eur. J. Pure Appl. Math, 10 (3) (2017), 506-515 509

Lemma 3. For n ≥ 0, the sum
n∑

i=0

(
n
i

)
ckivki = hkn satisfies the recurrence relation

hk(n+2) =
(
ckvk + 2

)
hk(n+1) −

(
c2k(−q)k + ckvk + 1

)
hkn,

where initial conditions h0 = 2, hk = 2 + ckvk.

Lemma 4. The generating function V (z) of
n∑

i=0

(
n
i

)
ckivki = hkn is given by

V (z) =
zk
(
hk − 2

(
ckvk + 2

))
+ 2

1− (ckvk + 2) zk + (c2k(−q)k + ckvk + 1) z2k
.

If ckαk + 1 and ckβk + 1 are the roots of equation

x2 −
(
ckvk + 2

)
x+

(
c2k(−q)k + ckvk + 1

)
= 0,

Binet formulae of sequences {gkn} and {hkn} are

gkn =

(
ckαk + 1

)n − (ckβk + 1
)n

α− β
and hkn =

(
ckαk + 1

)n
+
(
ckβk + 1

)n
,

respectively. It is clear that

g−kn = −
(
c2k(−q)k + ckvk + 1

)−n
gkn, g2kn = gknhkn (2)

and h−kn =
(
(−q)kc2k + ckvk + 1

)−n
hkn.

Now, we define a 2× 2 matrix A and then we give some new results for the sequences
{gkn} and {hkn} by matrix methods.

Consider the 2× 2 matrix A as follows:

A =

[
ckvk + 2 −

(
c2k(−q)k + ckvk + 1

)
1 0

]
.

The eigenvalues of the matrix A are

λ1 = ckαk + 1, λ2 = ckβk + 1.

Also λ1, λ2 are distinct. Let V be the 2× 2 matrix defined as follows:

V =

[
ckαk + 1 ckβk + 1

1 1

]
.

One can easily verify that
AV = V D1,



N. Ömür, C. D. Şener / Eur. J. Pure Appl. Math, 10 (3) (2017), 506-515 510

where D1 = diag (λ1, λ2) . Since detV 6= 0, the matrix V invertible. So, we write

D1 = V −1AV.

Thus, the matrix A is similar to the diagonal matrix D1. We obtain

An = V Dn
1V
−1

=
1

ckuk

 gk(n+1) −
(
c2k (−q)k + ckvk + 1

)
gkn

gkn −
(
c2k (−q)k + ckvk + 1

)
gk(n−1)

 .
Clearly, the matrix An satisfies the recurrence relation: for n > 0,

An+1 =
(
ckvk + 2

)
An −

(
c2k(−q)k + ckvk + 1

)
An−1,

where initial conditions A0 = 0, A1 = A.
Also by matrix methods, it is clearly that

An

[
gk
g0

]
=

[
gk(n+1)

gkn

]
. (3)

For n ≥ 0, if we consider the fact that det(An) = (detA)n, then we obtain the Cassini
identity

g2kn − gk(n+1)gk(n−1) =
(
c2k (−q)k + ckvk + 1

)n−1
g2k.

For example, for k = c = p = q = 1, we write F 2
2n−F2n+1F2n−1 = −1( see page 74, [12]).

Similarly

A−n =
1

ckuk

(
c2k (−q)k + ckvk + 1

)n ×
[
−
(
c2k (−q)k + ckvk + 1

)
gk(n−1)

(
c2k (−q)k + ckvk + 1

)
gkn

−gkn gk(n+1)

]

and

A−n
[
gk
g0

]
=

[
gk(−n+1)

g−kn

]
. (4)

By considering sequence {hkn}, we write the simple relation between the vector of
sequence {hkn} and generating matrix of sequence {gkn} :

[
hk(n+1)

hkn

]
=

1

ckuk

 gk(n+1) −
(
c2k (−q)k + ckvk + 1

)
gkn

gkn −
(
c2k (−q)k + ckvk + 1

)
gk(n−1)

[ ckvk + 2
2

]
.



N. Ömür, C. D. Şener / Eur. J. Pure Appl. Math, 10 (3) (2017), 506-515 511

Theorem 1. For all n,m ∈ Z, we have

ckukgk(n+m) = gkngk(m+1) −
(
c2k (−q)k + ckvk + 1

)
gk(n−1)gkm. (5)

Proof. After some simplications,(2, 1)−entries of AnAm = An+m give the conclusion.

For example, when n = m in (5), we have

ckukg2kn = gkngk(n+1) −
(
c2k (−q)k + ckvk + 1

)
gkngk(n−1),

or
ckukhkn = gk(n+1) −

(
c2k (−q)k + ckvk + 1

)
gk(n−1).

Theorem 2. For all n ∈ Z, we have

ckuk
(
gk(2n+1) + gk(2n−1)

)
= g2k(n+1) −

(
c2k (−q)k + ckvk

)
g2kn −

(
c2k (−q)k + ckvk + 1

)
g2k(n−1),

ckuk
(
gk(2n+1) − gk(2n−1)

)
= g2k(n+1) −

(
c2k (−q)k + ckvk + 2

)
g2kn +

(
c2k (−q)k + ckvk + 1

)
g2k(n−1).

Proof. Considering the (1, 1) and (2, 2)−entries of the matrix equation A2n = (An)2,
we have

g2k(n+1) −
(
c2k (−q)k + ckvk + 1

)
g2kn = ckukgk(2n+1), (6)

g2kn −
(
c2k (−q)k + ckvk + 1

)
g2k(n−1) = ckukgk(2n−1). (7)

By adding and substracting of (6) and (7) side by side, we have the conclusion.

Theorem 3. For n > 0, we have

gkn =
n∑

t=0

(
n

t

)(
ckvk + 2

)t
(−1)n−t

(
c2k(−q)k + ckvk + 1

)−n+t
gk(n−t).

Proof. From the matrix relation, we write

An =
((
ckvk + 2

)
I −

(
c2k(−q)k + ckvk + 1

)
A−1

)n
=

n∑
t=0

(
n

t

)(
ckvk + 2

)t (
c2k(−q)k+1 − ckvk − 1

)n−t
A−(n−t). (8)

Then equating (2, 1)−entries of the equality (8), we get

gkn =

n∑
t=0

(
n

t

)(
ckvk + 2

)t
(−1)n−t

(
c2k(−q)k + ckvk + 1

)−n+t
gk(n−t).



N. Ömür, C. D. Şener / Eur. J. Pure Appl. Math, 10 (3) (2017), 506-515 512

Theorem 4. For n > 0, we have

2n∑
i=0

(
2n

i

)(
c2k(−q)k + ckvk + 1

)2n−i
gk(2i+1) =

(
ckvk + 2

)2n
gk(2n+1),

2n∑
i=0

(
2n

i

)(
c2k(−q)k + ckvk + 1

)2n−i
g2ki =

(
ckvk + 2

)2n
g2kn.

Proof. The matrix A2 is[ (
ckvk + 2

)2
+ qk(−1)k+1c2k − ckvk − 1

(
ckvk + 2

) (
qk(−1)k+1c2k − ckvk − 1

)
ckvk + 2 qk(−1)k+1c2k − ckvk − 1

]
and the characteristic equation for A2 is

λ
(
ckvk + 2

)2
=
(
λ+

(
c2k(−q)k + ckvk + 1

))2
. (9)

From the Caley-Hamilton Theorem for A2, we have

A2
(
ckvk + 2

)2
=
(
A2 +

(
c2k(−q)k + ckvk + 1

)
I
)2
.

Thus

A2n
(
ckvk + 2

)2n
=
(
A2 +

(
c2k(−q)k + ckvk + 1

)
I
)2n

.

By Binomial Theorem and (3), we have

2n∑
i=0

(
2n

i

)(
c2k(−q)k + ckvk + 1

)2n−i
A2i = A2n

(
ckvk + 2

)2n
2n∑
i=0

(
2n

i

)(
c2k(−q)k + ckvk + 1

)2n−i [ gk(2i+1)

g2ki

]
=

(
ckvk + 2

)2n [ gk(2n+1)

g2kn

]
,

as claimed.

Theorem 5. For n > 0, we have

2n∑
i=0

(
2n

i

)
(−1)i

(
c2k(−q)k + ckvk + 1

)2n−i
gk(2i+1) = c2kn

(
v2k + 4qk (−1)k+1

)n
gk(2n+1),

2n∑
i=0

(
2n

i

)
(−1)i

(
c2k(−q)k + ckvk + 1

)2n−i
g2ki = c2kn

(
v2k + 4qk (−1)k+1

)n
g2kn.

Proof. Writing 4λ
(
qk(−1)k+1c2k − ckvk − 1

)
to each side of (3), we write

λc2k
(
v2k − 4 (−q)k

)2
=
(
λ+

(
qk(−1)k+1c2k − ckvk − 1

))2
. (10)

Similarly, using (10), as the proof of Theorem 1, the proof is completed.



N. Ömür, C. D. Şener / Eur. J. Pure Appl. Math, 10 (3) (2017), 506-515 513

Theorem 6. For n > 0, we have

2n∑
i=0

(
2n

i

)(
c2k(−q)k + ckvk + 1

)i
gk(−2i+1) =

(
ckvk + 2

)2n
gk(−2n+1),

2n∑
i=0

(
2n

i

)(
c2k(−q)k + ckvk + 1

)i
g−2ki =

(
ckvk + 2

)2n
g−2kn.

Proof. The characteristic equation for A−2 is(
λ
(
c2k(−q)k + ckvk + 1

)
+ 1
)2

= λ
(
ckvk + 2

)2
.

From the Caley-Hamilton Theorem for A−2, we have(
A−2

(
c2k(−q)k + ckvk + 1

)
+ 1
)2

=
(
ckvk + 2

)2
A−2.

Thus (
A−2

(
c2k(−q)k + ckvk + 1

)
+ 1
)2n

=
(
ckvk + 2

)2n
A−2n.

By Binomial Theorem and (4), we have

2n∑
i=0

(
2n

i

)(
(−q)kc2k + ckvk + 1

)i
A−2i =

(
ckvk + 2

)2n
A−2n

2n∑
i=0

(
2n

i

)(
c2k(−q)k + ckvk + 1

)i [ gk(−2i+1)

g−2ki

]
=

(
ckvk + 2

)2n [ gk(−2n+1)

g−2kn

]
.

Thus, the proof is completed.

Let C be an arbitrary 2× 2 matrix, T and D denote the trace and determinant of C,
respectively. For the distinct eigenvalues a and b of matrix C, the following result can be
found in [5, 10]:

Lemma 5.

zn :=
an − bn

a− b
=

1

2n−1

b(n−1)/2c∑
i=0

(
n

2i+ 1

)
Tn−2i−1(T 2 − 4D)i,

then Cn = znC − zn−1DI2, where I2 is the identity matrix of order 2.

As a consequence of Lemma 5, we obtain that

2n−1gkn = ukc
k

b(n−1)/2c∑
i=0

(
n

2i+ 1

)(
ckvk + 2

)n−2i−1
(c2kv2k − 4c2k (−q)k)i.



REFERENCES 514

Let w be a complex number such that w2 + Tw +D 6= 0, w 6= 0. For a positive integer n,

Cn =

(
wD

w2 + Tw +D

)n 2n∑
t=0

t∑
i=0

(
n

i

)(
n

t− i

)(
D

w2

)i (w
D

)t
Ct. (11)

Therefore we get the following result of equality (11).

Theorem 7. For n > 0 and any complex number w different from 0, ckαk+1 and ckβk+1,

gkn =

 w
(
c2k (−q)k + ckvk + 1

)
w2 + (ckvk + 2)w + (c2k(−q)k + ckvk + 1)

n

×

2n∑
t=0

t∑
i=0

(
n

i

)(
n

t− i

)(
qk(−1)kc2k + ckvk + 1

)i−t
wt−2igkt.

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