Evaluation of various partial sums of Gaussian qbinomial sums
Evaluation of various partial sums of Gaussian q binomial sums
Emrah Kılıç
Mathematics Subject Classification
We present three new sets of weighted partial sums of the Gaussian qbinomial coefficients. To prove the claimed results, we will use qanalysis, Rothe's formula and a qversion of the celebrated algorithm of Zeilberger. Finally we give some applications of our results to generalized Fibonomial sums. r 2 − k

1 Introduction
The authors of [
2
] present some results about the partial sums of rows of Pascal’s triangle as well as its
alternating analogues. As alternating partial sum of Pascal’s triangle, from [
2
] we have
k≤m
r
(−1)k = (−1)m
r − 1
m
The authors note that there is no closed form for the partial sum of a row of Pascal’s triangle, S (m, n) =
nk . However, they give a curious example for the partial sum of the row elements multiplied by their
diskt≤anmce from the center:
k≤m
r
=
m + 1
2
r
m + 1
Ollerton [
11
] considered the same partial sum of a row of Pascal’s triangle S (m, n) and developed formulae
for the sums by use of generating functions.
We recall a different result from [
6
]: for any nonnegative integer t ,
k≤t
2n
k
k
1 − n
=
.
These kinds of partial sums, as well as certain weighted sums (both alternating, and non alternating) are
very rare in the current literature. Calkin [
1
] proved the following curious identity:
Hirschhorn [
5
] established the following two identities on sums of powers of binomial partial sums:
n
k
k=0 j =0
n
j
n ⎛
⎝
k=0
k
j =0
n ⎞ 3
j ⎠
= n · 2n−1 + 2n
and
Recently, He [
4
] gave q analogues of the results of Hirschhorn as well as a new version of another result.
For example, from [
4
], we recall that
,
where the q Pochhammer symbol is (x ; q )n = (1 − x )(1 − x q ) . . . (1 − x q n−1) and the Gaussian q binomial
coefficients are
Meanwhile, Guo et al. [
3
] gave some partial sums including q binomial coefficients, for example
For later use, we recall that one version of the Cauchy binomial theorem is given by
n
k q = (q ; q )k (q ; q )n−k
(q ; q )n
.
n
lim
q→1 k q
=
n
k
.
⎛
⎝
k
j =0
n
k=0
2n
k=0
(−1)k
2n
j
q ⎠
⎞ 2
=
2n
k=0
2n
k
q
2n
k=0
2n
2k q
.
q
k+1
2
n
k q
x k =
n
k=1
(1 + x q k ),
n
k=0
k
(−1)k q 2
n
k q
x k
= (x ; q )n =
(1 − x q k ).
n−1
k=0
Note that
and Rothe’s formula is
Nowadays, there is increasing interest in deriving q analogues of certain combinatorial statements. This
includes, for example, q analogues of certain identities involving harmonic numbers, we refer to [
9, 10
].
and
and
Recently, Kılıç and Prodinger [
7
] computed half Gaussian q binomial sums with certain weight functions.
They also gave applications of their results as the generalized Fibonomial sum formulas. For example, they
showed that for n ≥ 1 such that 2n − 1 ≥ r
n
k=0
More recently, Kılıç and Prodinger [
8
] computed three types of sums involving products of the Gaussian
q binomial coefficients. They are of the following forms: for any real number a
In this section, we will examine each set in separate subsections. We start with the first set of weighted partial
sums.
2.1 First kind of weighted partial sums
We start with the following lemma to use in proving our first set of results.
Lemma 2.1 (i) For odd n,
(ii) For even n,
n
k=0
n
k q
n
k=0
n
k q
q 21 k(k−n)ik(n−k) (−1)k = 0.
q 21 k(k−n−1)ik(n+k)+k (1 − q k ) = 0.
Proof We only give a proof for the first case (i). The second case (ii) is similarly done. For odd n, we write
k=0
Here we consider two subcases: first, if n is even,
= 2i1−nq−(n+21) n (1 + q2k)[(−1)n − 1],
k=1
Apart from the constant factor, we consider the LHS of the claim and write
By Lemma 2.1 (ii), we have
k=m
k=0
=
k=m
2n+1 2n + 2m + 4 (1 − q2n+m+4−k)(−1)(2)+kn q 21k(k−3)−kn
k
k + m q
×
2n+m+1 2n + 2m + 4 (1 − q2n+2m+4−k)(−1)k(m+n)−(k2) q 21k(k−2(m+n)−3).
k q
2n+2m+4 2n + 2m + 4 (1 − q2n+2m+4−k)(−1)k(m+n)−(2k) q 21k(k−2(m+n)−3) = 0.
k q
−
2n+m+1 2n + 2m + 4 (1 − q2n+2m+4−k)(−1)k(m+n)−(k2) q 21k(k−2(m+n)−3)
k q
k=0
m−1 2n + 2m + 4 (1 − q2n+2m+4−k)(−1)k(m+n)−(k2) q 21k(k−2(m+n)−3)
k q
k=0
(−1)(k+1)(m+n)−(k +21) q− 21 (k−1)(2m−k+2n+4) 1 − qk .
k=0
+
k=0
m+2 2n + 2m + 4
k
After further rearrangements, this equals
which, by telescoping, equals
k=2n+m+2 q
× (−1)k(m+n)−(k2) q 21 k(k−2(m+n)−3),
+
which, after some rearrangements, equals
1 − q2n+2m+4
m−1 2n + 2m + 3
k
q
(−1)k(m+n)− 21 k(k−1) q 21 k(k−2m−2n−3)
(−1)(k+1)(m+n)−(k +21) q− 21 (k−1)(2m−k+2n+4)
+
+
−
+
×
k=0
k=0
m+2 2n + 2m + 3
k − 1
k=0
k=0
m+1 2n + 2m + 3
k
k=0
k=0
m+1 2n + 2m + 3
k
q
q
q
= (1 − q2n+2m+4) m−1 2n + 2m + 3
k
(−1)k(m+n)−(2k) q 21 k(k−2m−2n−3)
= 1 − q2n+2m+4
m−1 2n + 2m + 3
k
(−1)k(m+n)−(2k) q 21 k(k−2m−2n−3)
(−1)k(m+n)−(k2)+1 q 21 k(k−2m−2n−3)
(−1)k(m+n)−(k2) q 21 k(k−2m−2n−3) ,
q
q
− 1 − q2n+2m+4
2n + 2m + 3
m + 1
q
(−1) 21 (m+2n)(m+1) q− 21 (m+1)(m+2n+2)
2n + 2m + 3
m
(−1) 21 m(m+2n+1) q− 21 m(m+2n+3)
q
= − 1 − q2n+2m+4 (−1) 21 m(m+2n+1) q− 21 m(m+2n+3)
2n + 2m + 3 (−1)n q−(n+1) + 2n + 2m + 3
m + 1 q m
q
Thus, we get as claimed. We present the following results without proof.
Theorem 2.3 For n ≥ 0 and m ≥ 1,
= −(−1)21m(m+2n+1) q−21(m+1)(m+2n+2) 2n + 2m + 4
m + 1 q
1 + qn+1 1 − qm+n+2 if n is even,
× − 1 − qn+1 1 + qm+n+2 if n is odd.
2n+1 2n + 2m + 4
k=0
Theorem 2.4 For n,m ≥ 0,
2n 2n + 2m − 1 q(k+21)−kn (−1)kn−(k2) = (−1)n qn 2n + 2m − 1 ,
k=0 k + m q m − 1 q
2n+1 2n + 2m + 3 q 21k(k−2n−3) (−1)(k2)+kn
k=0 k + m q
= (−1)n q−n−1 2n +m2+m1+ 4 q (11+−qqnn++11)//(11−+qqnn++mm++22) iiff nn iiss eovdedn,.
2.2 Second kind of weighted partial sums
Now we continue with our second set of sums. Before that, we need the following lemma which could be
proven similar to Lemma 2.1.
Lemma 2.5 (i) If n is odd,
(ii) If n is even,
We start with the following result.
Theorem 2.6 For n, m ≥ 0,
4n+1 4n +k+2mm+ 3 q (−1)k q 21 k(k−4n−3) = −q−2n−1 (1 − q2n(+11−)(q1m++q1)2n+m+2) 4n + m2m + 3 q.
Proof Consider the LHS of the claim, by taking k−m instead of k, we note
k=0
4n+1 4n + 2m + 3
k + m
q
and by Lemma 2.5 (i),
Thus, we obtain
which, by some rearrangements, equals
k=0
By telescoping, this equals
which equals
− (−1)m q− 21 (m+1)(m+4n+2) − 4n + 2m + 3
m + 1
q
+
4n + 2m + 3 q2n+1
m q
= − (−1)m q− 21 (m+1)(m+4n+2)
(q)4n+2m+3
(q)m+1 (q)4n+m+3
× (−(1 − q4n+m+3) + q2n+1(1 − qm+1)),
Thus, we derive
Using similar techniques as in the above proof, we obtain the following identities, for all n,m ≥ 0,
4n+1 4n + 2m − 1 (−1)k q(k+21)−2kn = q2n (1 − q2n+1)(1 + qm+2n) 4n + 2m − 1 ,
k=0 k + m q 1 − qm−1 m − 2 q
4n 4n + 2m − 1 q(k+21)−2kn (−1)k = q2n 4n + 2m − 1 ,
k=0 k + m q m − 1 q
q−21k(4n−k+3) (−1)k (1 − q4n+4+m−k) = (1 − q4n+4+2m) 4n + 2m + 3 ,
m q
(1)
4n+2 4n + 2m + 4
k=0
4n 4n + 2m
k=0 k + m q
2.3 Third kind of weighted partial sums
We start with the following lemma for later use.
Lemma 2.7 For m > 0 and integer r,
q 21k(k−4n+1) (−1)k (1 − q4n+m−k)
m+r 4n + 2m + 2
k=0
k
q
(−1)k q 21k(k−4n−2m−1)(1 + q2n+m+1−k)
= (−1)m+rq−21(m+r)(m+4n−r+1)(1 + q1+m+2n) 4n + 2m + 1 .
m + r q
Proof Denote the lefthand side of this identity by S[m,r]. Using the celebrated algorithm of Zeilberger in
Mathematica, we obtain
which is easily seen to be the same as the righthand side of the desired identity.
Theorem 2.8 For n ≥ 0 and m > 0,
4n+3 4n + 2m + 2 q 21k(k−4n−3) (−1)k (1 − qk)
4n+3 4n + 2m + 2 q 21k(k−4n−3) (−1)k (1 − qk)
k=0
which, by Lemma 2.7 with the case r = −2, equals
4n+3 4n + 2m + 2 q 21k(k−4n−3) (−1)k (1 − qk) = −q2n+1(1 − q2n+2)(1 − qm) 4n + 2m + 2 ,
k=0 k + m q (1 − qm+2n+1) m − 1 q
We present the following results without proof. The proofs use Lemma 2.7 and are similar to that of
Theorem 2.8.
Theorem 2.9 For n ≥ 0 and m ≥ 1,
4n+2 4n + 2m + 2 q 21k(k−4n−3) (−1)k (1 − qk) = (1 − q2n+1)(1 + q1+m+2n) 4n + 2m + 1 .
k=0 k + m q m − 1 q
Theorem 2.10 For n,m ≥ 0,
4n+1 4n + 2m + 2 (−1)k q 21k(k−4n−3)(1 − qk) = −q−2n−1(1 + q2n+m+1)(1 − q2n+1) 4n + 2m + 1 ,
k + m q m q
k=0
k=0
In this section, we will give some applications of our results to the generalized Fibonomial coefficients. Define
the nondegenerate secondorder linear sequences {Un} and {Vn} by, for n > 1
The Binet formulas of {Un} and {Vn} are
where α,β = (p ± p2 + 4)/2.
For n ≥ k ≥ 1, define the generalized Fibonomial coefficient by
Un = pUn−1 + Un−2 , U0 = 0, U1 = 1,
Vn = pVn−1 + Vn−2 , V0 = 2, V1 = p.
αn − βn
Un = α − β and Vn = αn + βn ,
n
k U := (U1U2 ...Uk) · (U1U2 ...Un−k)
with
n
0 U =
When p = 1, we obtain the usual Fibonomial coefficients, denoted by nk F .
The link between the generalized Fibonomial and Gaussian q binomial coefficients is
n
k U
= αn−k
n
k q
with
q = −α−2.
By taking q = β/α, the Binet formulae are reduced to the following forms:
Un = αn−1
1 − q n
1 − q
and
Vn = αn (1 + q n ),
where i =
√
,
,
(−1) 21 k(k−1) Uk = U2n+1 V1+m+2n
4n + 2m + 1
m − 1
U
.
As a showcase we will prove the second identity just above. First we convert the second identity into
q form. Thus, it takes the form
q
q
q
or
or
4n+2
k=0
4n + 2m + 3
k + m
q
4n+2
k=0
4n + 2m + 3
k + m
q
α(k+m)(4n+2m+3−(k+m)) (−1) 21 k(k−1)
=
4n + 2m + 3
m
αm(4n+2m+3−m)
α(k+m)(m−k+4n+3) (−1) 21 k(k−1)
=
4n + 2m + 3
m
αm(4n+m+3)
4n+2
k=0
4n + 2m + 3
k + m
q
(−1)k q − 21 k(4n−k+3)
=
4n + 2m + 3
m
which was already given as the identity (1).
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