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CERTAIN BINOMIAL SUMS WITH RECURSIVE COEFFICIENTS
EMRAH KILIC AND EUGEN J. IONASCU
Abstract. In this short note, we establish some identities containing sums of binomials
with coefficients satisfying third order linear recursive relations. As a result and in particular,
we obtain general forms of earlier identities involving binomial coefficients and Fibonacci
type sequences.
1. Introduction
There are many types of identities containing sums of certain functions of binomial coefficients and Fibonacci, Lucas, or Pell numbers. Let us give a few examples of such identities
(see [2, 5]):
n
X
¡n¢
k
Fk = F2n ,
k=0
n
X
¡n¢
k
2k Fk = F3n ,
k=0
n ≥ m,
n
X
2n
X
¡2n¢
F2k = 5n F2n
k
(1.1)
k=0
¡ ¢
(−1) nk Fn+k−m = Fn−m ,
k
2n
X
¡2n¢ 2
F2k = 5n−1 L2n ,
k
(1.2)
k=0
k=0
2n
X
¡2n¢
L2k = 5n L2n ,
k
2n
X
k=0
k=0
(−1)k
¡2n¢ k−1
2 Lk = 5n ,
k
(1.3)
where Fn and Ln stand as usual for the nth Fibonacci and respectively the nth Lucas
number. We remind the reader that F0 = 0, F1 = 1, L0 = 2, L1 = 1 and Fn+1 = Fn + Fn−1 ,
Ln+1 = Ln + Ln−1 for n ∈ N. As a more sophisticated example the following identity was
asked by Hoggatt as an advanced problem in [4]:
n
X
¡ n¢
F4mk = Ln2m F2mn .
(1.4)
k
k=0
Generalizations of the above identities appeared in [1, 3]. For instance, if u, v and r are
integers, uv(u − v) 6= 0, then
n
X
¡ ¢ n−k k
n
Fv Fun+r =
(−1)(n−k)u nk Fv−u
Fu Fvk+r ,
k=0
which generalizes the identities (1.1–1.2). Similar identities to (1.1–1.3) for the Pell numbers
can be derived. Our interest here is for identities in which only half of the binomial coefficients
are used. Three such identities are
MAY 2010
161
THE FIBONACCI QUARTERLY
n
X
¡
k=0
n
X
2n
n+k
¢
Fk2 = 5n−1 , n ∈ N,
¡ 2n ¢
n+k
k=0
n
X
¡ 2n ¢
n+k
L2k = 5n + 2
¡2n¢
n
(1.5)
, n ∈ N,
(1.6)
Pk2 = 8n−1 , n ∈ N,
(1.7)
k=0
where Pn is the nth Pell number (P0 = 0, P1 = 1, and Pn+1 = 2Pn + Pn−1 , n ∈ N). We are
also going to work with generalized Fibonacci {un } and Lucas {vn } sequences defined by
un+1 = pun + un−1 , vn+1 = pvn + vn−1 n ∈ N,
(1.8)
where u0 = 0, u1 = 1 or v0 = 2, v1 = p for any complex number p.
One can derive easily the Binet formulas for {un } and {vn }:
un =
αn − β n
and vn = αn + β n , n ∈ N ∪ {0},
α−β
³
´
³
´
p
p
2
2
where α = p + p + 4 /2 and β = p − p + 4 /2, with the principal branch of the
√
√ θ
square root reiθ = re 2 , θ ∈ (−π, π], r ≥ 0. From these formulas, we can easily get the
following identity:
¡
¢
vn+m − (−1)m vn−m = p2 + 4 un um ,
m, n ∈ N ∪ {0}.
(1.9)
In this note we will establish generalizations of the formulas (1.5), (1.6), (1.7) and a version
of (1.9) with combinatorial coefficients involved. Our techniques are purely computational
and very much in line with the standard ones used to show (1.1)–(1.3). One way of generalizing the identities in (1.5)–(1.7) is to use different powers for the recursive function
term:
n
X
¡
2n
n+k
¢
k=0
n
X
¡
2n
n+k
¢
k=0
n
X
¡
k=0
2n
n+k
¢
1 ¡ 2n
3 − 4 (−1)n + 3 × 22n , n ∈ N,
25
µ ¶
2n
2n
2n
= 3 +3×2 +8
+ 4 (−1)n , n ∈ N,
n
Fk4 =
¢
L4k
Pk4 =
¢
1 ¡ 2n
6 − 22n+2 − 4 (−1)n + 3 × 22n , n ∈ N.
64
(1.10)
(1.11)
(1.12)
An interesting question at this point is whether or not one can arrange so that for some
Fibonacci type recurrent sequence, powers of any positive integer could be represented as in
(1.5) and (1.7). We will address this question in the last section of this note.
162
VOLUME 48, NUMBER 2
CERTAIN BINOMIAL SUMS WITH RECURSIVE COEFFICIENTS
2. Half of the Binomial Formula
In this section we build up the main ingredients for our calculations. Let us define the
function f of a ∈ C \ {0} and n ∈ N:
¶
n µ
X
¡ k
¢
2n
f (n, a) =
a + a−k .
n+k
k=0
We have the following lemma.
Lemma 2.1. For every non-negative integer n and a ∈ C \ {0} the function f satisfies
¡ ¢
1
.
f (n, a) = n (a + 1)2n + 2n
n
a
Proof. We can write
¡2n¢ 2n
¡ ¢ n ¡ 2n ¢ n+1
a
a
+
·
·
·
+
an f (n, a) = 2n
a
+
n
n+1
2n
¡2n¢ 0
¡2n¢ n ¡ 2n ¢ n−1
+ · · · + 2n a
+ n a + n+1 a
¡¡2n¢ n ¡ 2n ¢ n+1
¡2n¢ 2n ¢
a
=
a
+
a
+
·
·
·
+
n+1
2n
n
¡¡ 2n ¢ n−1
¡2n¢ 0 ¢ ¡2n¢ n
+ n−1 a
+ ··· + 0 a + n a
¡2n¢ n
2n
= (a + 1) + n a .
Hence the identity claimed follows by dividing by an .
¤
An observation here is necessary. We formulate this as a proposition.
Proposition 2.2. For every non-negative integer n and a ∈ C \ {0} the following formulas
yield true:
¶
µ
¶
n µ
n
X
2n
2n
1 ¡2n¢ X
1 ¡2n¢
2n−1
k
=2
+
(−1)
=
,
,
n+k
n+k
2 n
2 n
k=0
k=0
¶2n
µ
¡ ¢
¡2n¢
1
r
r 2n−1
2
n
+ 2n
.
f (n, (−1) ) = (1 + (−1) )2
+ n , f (n, −a ) = (−1) a −
n
a
Proof. To obtain the first two identities we set a = 1 and then a = −1. The last claim is
obtained by substituting a with −a in Lemma 2.1.
¤
3. Proof of the Claimed Identities
We will work with the generalized Fibonacci type sequences {uk } and {vk } defined in the
introduction. The formulas (1.5), (1.10), (1.7), and (1.12) are contained in the next theorem.
Theorem 3.1. For n ∈ N ∪ {0} and r ∈ N, and {uk } and {vk } defined as before by (1.8),
we have
õ ¶
!

µ ¶
r−1
X

2r
2r
1

2n

22n−2 +
(−1)i(n+1)
vr−i
, if r is even;

2 + 4)r

(p
r
i

µ
¶
n

i=0
X
2n
=
u2r
à r−1
!

n+k k
µ ¶

k=0

X
¡
¢

2r
n−r
2


(−1)i(n+1)
u2n
,
if r is odd.
 p +4
r−i
i
i=0
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163
THE FIBONACCI QUARTERLY
Proof. We expand first uk to the power 2r using the binomial formula:
#
µ ¶
¶
¶ "X
n µ
n µ
2r
X
X
2n
2r
2n
1
(−1)i
α(2r−i)k β ik .
u2r
k =
2r
n+k
i
(α − β) k=0 n + k
i=0
k=0
¡ ¢ ¡ 2r ¢
Taking into account that αk β k = (−1)k and the facts that 2ri = 2r−i
, (−1)i = (−1)2r−i ,
for i = 0, 1, 2, . . . , 2r, we can turn the above into
¶
¶·
µ ¶
n µ
n µ
X
X
2n
1
2n
2r
r(1+k) 2r
uk = 2
(−1)
r
n
+
k
(p
+
4)
n
+
k
r
k=0
k=0
¸
µ ¶
r−1
X
2r
i+ik
2(r−i)k
−2(r−i)k
(α
+α
) .
+
(−1)
i
i=0
Commuting the two summations and using Lemma 2.1 (Proposition 2.2) the calculation
can be continued to
¶
n µ
X
2n
u2r
n+k k
k=0
¶
µ ¶
µ
µ ¶
r−1
X
2r
(−1)r 2r
1
i 2(r−i)
r
i
f (n, (−1) α
)
f (n, (−1) ) +
(−1)
= 2
i
r
(p + 4)r
2
i=0
µ
µ ¶
1
(−1)r 2r
= 2
f (n, (−1)r )
r
(p + 4)r
2
µ ¶h
r−1
X
¡2n¢
¡
¢ i
i 2r
in
r−i r−i
i r−i 2n
+
(−1)
(−1)
β
+
(−1)
α
+
(−1)
n
i
i=0
!
õ ¶

µ ¶
r−1
X

1
2r 2n
2r 2n−1


vr−i , if r is even;
2
+
(−1)i(n+1)

r
2

i
r

 (p + 4)
i=0
=
à r−1
!

µ ¶


X
¡
¢

2r
n−r
2


u2n
if r is odd.
(−1)i(n+1)
 p +4
r−i ,
i
i=0
¤
The similar result to Theorem 3.1 but for {vn } is stated next.
Theorem 3.2. For n ∈ N ∪ {0} and r ∈ N, and {uk } and {vk } defined as before by (1.8),
we have
µ ¶
µ ¶
r−1
¡2n¢ X

2r 2n−1
2r 2n

2r−1
in

2
+2
+
(−1)
vr−i , if r is even;

n

r
i

µ
¶
n

i=0
X
2n
vk2r =
à r−1
!

n+k
µ ¶

k=0

X
¡
¢
¡
¢

2r
n
2r−1 2n


+ p2 + 4
(−1)in
u2n
2
r−i , if r is odd.
n
i
i=0
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VOLUME 48, NUMBER 2
CERTAIN BINOMIAL SUMS WITH RECURSIVE COEFFICIENTS
Proof. In this case after we expand vk to the power 2r, we get:
#
¶
¶ "X
n µ
n µ
2r µ ¶
X
X
2r
2n
2n
vk2r =
α(2r−i)k β ik .
i
n
+
k
n
+
k
i=0
k=0
k=0
¡ ¢
¡ 2r ¢
k k
k
Again using that α β = (−1) and the facts that 2ri = 2r−i
, (−1)i = (−1)2r−i , for
i = 0, 1, 2, . . . , 2r, we can turn the above into
#
¶"
µ ¶ X
µ ¶
¶
n µ
r−1
n µ
X
X
2n
2r
2n
2r
(−1)rk
+
(−1)ik
vk2r =
(α2(r−i)k + α−2(r−i)k ) .
n
+
k
r
n
+
k
i
i=0
k=0
k=0
Commuting the two summations and using Proposition 2.2 the above can be continued
into
à µ ¶
!
¶
n µ
r−1 µ ¶
X
X
2n
1
2r
2r
vk2r =
f (n, (−1)r ) +
f (n, (−1)i α2(r−i) )
2
r
i
n
+
k
i=0
k=0
à µ ¶
!
µ
¶
r−1
i
X 2r h¡ ¢
¡
¢
1 2r
2n
2n
=
+ (−1)in (−1)r−i β r−i + (−1)i αr−i
f (n, (−1)r ) +
n
2 r
i
i=0
µ ¶
µ ¶
r−1

¡ ¢ X
2r 2n−1

2r−1 2n
in 2r
2n

+2
+
(−1)
vr−i
, if r is even;

n
 r 2
i
à r−1i=0
!
=
µ ¶
X

¡
¢
¡
¢
2r
n

2r−1 2n

+ p2 + 4
(−1)in
u2n

r−i , if r is odd.
2
n
i
i=0
¤
As another application of Lemma 2.1 to the second order recurrence {un }, we have the
following consequence.
Proposition 3.3. For n ≥ 0 and all even integers m ≥ 2 and t ≥ 0,
n
X
¡ 2n ¢
¡ 2
¢n ¡ 2n
¢
2n
(v
−
v
)
=
p
+
4
u
−
u
km
kt
m/2
t/2
n+k
(3.1)
k=0
where p is as before.
Proof. By Lemma 2.1 and the Binet formulas of the sequences {un } and {vn }, we write the
right side of (3.1) as
n
X
¡ 2n ¢ ¡ km
¢
km
kt
kt
α
+
β
−
α
−
β
n+k
k=0
¢
¡
= f (n, αm ) − f n, αt
¢2n
1 ¡
1
= mn (αm + 1)2n − tn αt + 1
α
α
¢2n
¢
¡
¡ m/2
2n
− αt/2 − β t/2
= α
− β m/2
¢
¡
2n
−
u
= (α − β)2n u2n
t/2
m/2
MAY 2010
165
THE FIBONACCI QUARTERLY
which completes the proof.
¤
One interesting consequence of Proposition 3.3 is the following.
Proposition 3.4. For n ≥ 0,
n
X
¡
2n
n+k
¢
¡
¢n
v2k = p2 + 4 .
k=0
Proof. We take m = 2, t = 0 in Proposition 3.3, and the proof follows.
¤
4. Representations of the Powers of Every Integer and Other Comments
To address the question we have raised in the introduction we notice that as a result of
Theorem 3.1 for r = 1 we obtain
¶
n µ
X
2n
u2k = (p2 + 4)n−1 , n ∈ N, p ∈ C.
n
+
k
k=0
Suppose we would like to have a power √
of 7 in the above equality, i.e. p2 + 4 = 7. This
can be accomplished if,√for instance, p = √
3. This turns {uk } and {vk } into sequences of
the form uk = ak + bk 3 and vk = ck + dk 3. Then the sequences {ak }, {bk }, {ck }, and
{dk } are uniquely determined by the recurrences
a0 = b0 = 0, a1 = 1, b1 = 0, an+1 = 3bn + an−1 , bn+1 = an + bn−1 , and
(4.1)
c0 = 2, d0 = 0, c1 = 1, d1 = 0,
(4.2)
cn+1 = 3dn + cn−1 , dn+1 = cn + dn−1 , n ∈ N.
Hence, for a generalized Fibonacci double sequence defined as in (4.1) we obtain a similar
identity to (1.5):
¶
n µ
X
2n
(a2k + 3b2k ) = 7n−1 , n ∈ N.
n+k
k=0
Let us observe that
n ¡
¢
P
2n
a b = 0 which implies that ak bk = 0 for every k ∈ N. This is a
n+k k k
k=0
little surprising since the two sequences {ak } and {bk } seem to be increasing. Theorem 3.1
for r = 4 gives
¶
n µ
X
¢
2n
1 ¡
70 × 22n−1 + 72n + 8(−1)n+1 42n + 28 × 32n + 56(−1)n+1 .
Fk8 =
625
n+k
k=0
Certain congruences can be derived by use of these identites. For example, for all n ∈ N,
32n − 4 (−1)n + 3 × 22n ≡ 0
(mod 25)
and
70 × 22n−1 + 72n + 8(−1)n+1 42n + 28 × 32n + 56(−1)n+1 ≡ 0
(mod 625).
Other such identities might be derived by the interested reader.
166
VOLUME 48, NUMBER 2
CERTAIN BINOMIAL SUMS WITH RECURSIVE COEFFICIENTS
References
[1] L. Carlitz, Some classes of Fibonacci sums, The Fibonacci Quarterly, 16.5 (1978), 411–426.
[2] R. A. Dunlap, The Golden Ratio and Fibonacci Numbers, World Scientific Publishing Co., River Edge,
NJ, 1997.
[3] P. Haukkanen, Formal power series for binomial sums of sequences of numbers, The Fibonacci Quarterly,
31.1 (1993), 28–31.
[4] V. E. Hoggatt, Advanced problem H-88, The Fibonacci Quarterly, 6.3 (1968), 253.
[5] S. Vajda, Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications, John Wiley
& Sons, Inc., New York, 1989.
MSC2010: 11B37
TOBB University of Economics and Technology, Mathematics Department 06560 Sögütözü,
Ankara, Turkey
E-mail address: [email protected]
Department of Mathematics, Columbus State University, 4225 University Avenue, Columbus, GA 31907
E-mail address: ionascu [email protected]
MAY 2010
167