# Some Finite Sums Involving Generalized Fibonacci and Lucas Numbers

Authors: E. Kılıç; N. Ömür; Y. T. Ulutaş
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

By considering Melham's sums (Melham, 2004), we compute various more general nonalternating sums, alternating sums, and sums that alternate according to(-1)(n+12) involving the generalized Fibonacci and Lucas numbers.


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## Body

## 1. Introduction

Leta, b, and p be assumed to be arbitrary nonzero complex numbers with p(p2+2)(p2+4)≠0. Define second-order linear recursion {Wn} by(1.1)Wn=pWn-1+Wn-2,
with W0=a,W1=b for all integers n. Since Δ=p2+4≠0, the roots α and β of x2-px-1=0 are distinct.Also define the sequence{Xn} via the terms of sequence {Wn} as Xn=Wn+1+Wn-1.The Binet formulas for the sequences{Wn} and {Xn} are(1.2)Wn=Aαn-Bβnα-β,Xn=Aαn+Bβn,
where A=b-aβ and B=b-aα.Fora=0,  b=1, we denote Wn=Un and so Xn=Vn, respectively. When p=1,  Un=Fn (nth Fibonacci number) and Vn=Ln (nth Lucas number).Inspired by the well-known identity(1.3)∑n=1jFn2=FjFj+1,
Clary and Hemenway [1] obtained factored closed-form expressions for all sums of the form ∑n=1jFmn3, where m is an integer. Motivated by the results in [1], Melham [2] computed all sums of the form ∑n=1j(-1)nFmn4 and ∑n=1j(-1)nLmn4. In [3], Melham computed various nonalternating sums, alternating sums, and sums that alternate according to (-1)(n+12) for sequences {Wn} and {Xn}. The author gathers his sums in three sets. Here we recall one example from each set for the reader's convenience:(1.4)∑n=ijWn={1pV(j-i+1)/2(W(j+i+1)/2+W(j+i-1)/2)ifj-i≡1(mod⁡4),1pU(j-i+1)/2(X(j+i+1)/2+X(j+i-1)/2)ifj-i≡3(mod⁡4),∑n=4i4j+3(-1)(n+12)W2n=pΔ-2U4j-4i+4X4j+4i+3,∑n=4i+24j+3(-1)(n+12)UnXn=pΔ-2V4j-4i+5W4j+4i+3+2W0.We refer to [4] for general expansion formulas for sums of powers of Fibonacci and Lucas numbers, as considered by Melham, as well as some extensions such that (1.5)∑k=0nF2k+δ2m+ϵ,∑k=0nL2k+δ2m+ϵ,
where δ,ϵ∈{0,1}.For alternating analogues of the results given by Prodinger, that is,(1.6)∑k=0n(-1)kF2k+δ2m+ϵ,∑k=0n(-1)kL2k+δ2m+ϵ,
we refer to [5].Hendel [6] gave the factorization theorem which exhibits factorizations of sums of the form ∑j=in+i-1Faj-b. The author also introduced a unified proof method based on formulae for the factorizations of Fq-d+Fq+d.In [7], Curtin et al. derived formulae for the shifted summations(1.7)∑j=0d-1Fn+jFm+j,∑j=0d-1Ln+jLm+j,∑j=0d-1Fn+jLm+j,
and the shifted convolutions(1.8)∑j=0d-1Fn+jFd-m-j,∑j=0d-1Ln+jLd-m-j,∑j=0d-1Fn+jLd-m-j
for positive integers d and arbitrary integers n and m.In this paper, our main purpose is to consider Melham's sums involving double products of terms of{Wn},{Xn},{Un}, and {Vn} given in [3] and then compute several more general nonalternating sums, alternating sums, and sums that alternate according to (-1)(n+12).


## 2. Certain Finite Sums of Double Products of Terms

In this section, we will investigate certain sums consisting of products of at most two terms of{Wn}: nonalternating sums, alternating sums and sums that alternate according to (-1)(n+12). From the Binet forms of {Wn} and {Xn}, we give the following lemma for further use without proof.Lemma 2.1.
Leta,   b, and  p be as in Section 1, and let  r=aW2-bW1. Then for all integers  k,
(2.1)b2U2k+2abU2k-1+a2U2k-2=WkXk,b2U2k+1+2abU2k+a2U2k-1=Wk+1Xk+(-1)kr,b2V2k+2abV2k-1+a2V2k-2=Xk2+(-1)k2r,b2V2k+1+2abV2k+a2V2k-1=XkXk+1+(-1)kpr.Theorem 2.2.
Fix integersc,d, and  m.(i)
Ifm is even, then for all integers j>i,(2.2)∑n=ijUmn+cWmn+d=Um(j-i+1)Xm(j+i)+c+dΔUm-(-1)c(j-i+1)Xd-cΔ.(i)
Ifm is odd, then for all integers j>i,(2.3)∑n=ijUmn+cWmn+d={Um(j-i+1)Wm(j+i)+c+dVmifj-i≡1(mod⁡2),Vm(j-i+1)Xm(j+i)+c+dΔVm-(-1)c+jXd-cΔifj-i≡0(mod⁡2).Proof.
Using the Binet formulas, we compute(2.4)∑n=ijUmn+cWmn+d=∑n=ij(αmn+c-βmn+cα-β)(Aαmn+d-Bβmn+dα-β)=1(α-β)2∑n=ij(Aα2mn+c+d+Bβ2mn+c+d)-(-1)mn+c(α-β)2(Aαd-c+Bβd-c)=1Δ∑n=ijX2mn+c+d-(-1)cΔXd-c∑n=ij(-1)mn.
Since Xn=Wn-1+Wn+1, we can obtain that for even m(2.5)∑n=ijX2mn+t=Um(j-i+1)Xm(j+i)+tUm.
The result follows.For example, wheni=2,m=3,a=0,b=c=p=1, and d=5, we obtain(2.6)∑n=2jF3n+1F3n+5=F3(j-1)F3j+44.Theorem 2.3.
Fix integersc,d, and  m. Let  S=∑n=ij(-1)nUmn+cWmn+d.(1)
Ifm is odd, then S equals
(2.7)S=(-1)jUm(j-i+1)Xm(j+i)+c+dΔUm-(-1)c(j-i+1)Xd-cΔ.(2)
Ifm is odd and the parities of i and j are the same, then S equals
(2.8)(-1)jVm(j-i+1)Xm(j+i)+c+dΔVm-((-1)j+(-1)i)(-1)cXd-c2Δ.(3)
Ifm is odd and the parities of i and j are the different, then S equals
(2.9)(-1)jUm(j-i+1)Wm(j+i)+c+dVm-((-1)j+(-1)i)(-1)cXd-c2.Proof.
Consider(2.10)∑n=ij(-1)nUmn+cWmn+d=∑n=ij(αmn+c-βmn+cα-β)(Aαmn+d-Bβmn+dα-β)=1(α-β)2∑n=ij(-1)n(Aα2mn+c+d+Bβ2mn+c+d)-(-1)mn+c(α-β)2(Aαd-c+Bβd-c)=1Δ∑n=ij(-1)nX2mn+c+d-1ΔXd-c∑n=ij(-1)(m+1)n+c.
SinceXn=Wn-1+Wn+1, for odd m, we find
(2.11)∑n=ij(-1)nX2mn+c=(-1)jUm(j-i+1)Xm(i+j)+cUm.
The result is now obtained by considering the values of ∑n=ij(-1)(m+1)n+c.Theorem 2.4.
Fix integersc,d, and  m. For all integers  j>i,
(2.12)∑n=4i+14j(-1)(n+12)Umn+cWmn+d=U4m(j-i)V2m{VmWs+mifmiseven,UmXs+mifmisodd,(2.13)∑n=4i4j+3(-1)(n+12)Umn+cWmn+d=U4m(j-i+1)V2m{UmXs+3mifmiseven,VmWs+3mifmisodd,∑n=4i+34j(-1)(n+12)Umn+cWmn+d={VmV2m(2(j-i)-1)Xs+3mΔV2m-2(-1)cXd-cΔifmiseven,UmV2m(2(j-i)-1)Ws+3mV2mifmisodd,∑n=4i+24j+3(-1)(n+12)Umn+cWmn+d={UmV2m(2(j-i)+1)Ws+5mV2mifmiseven,VmVm(4(j-i+1)-2)Xs+5mΔV2m+2(-1)cXd-cΔifmisodd,
where s=m(4(j+i))+c+d.Proof.
Consider(2.14)∑n=4i+14j(-1)(n+12)Umn+cWmn+d=∑n=4i+14j(-1)(n+12)(αmn+c-βmn+cα-β)(Aαmn+d-Bβmn+dα-β)=1(α-β)2∑n=4i+14j(-1)(n+12)(Aα2mn+c+d+Bβ2mn+c+d)-(-1)mn+c(α-β)2(Aαd-c+Bβd-c)=1Δ∑n=4i+14j(-1)(n+12)X2mn+c+d-1ΔXd-c(-1)c∑n=4i+14j(-1)(n+12).
Here we have that ∑n=4i+14j(-1)(n+12)=0 and, by Xn=Wn-1+Wn+1,
(2.15)∑n=4i+14j(-1)(n+12)X2mn+c+d=ΔVmU4m(j-i)Wm(4(j+i)+1)+c+dV2m,
for even m. Now formula (2.12) follows. The remaining formulas are proven in a similar manner.Notice that in (2.12)-(2.13), one limit of summation is even while the other is odd. Accordingly we have observed that each of (2.12)-(2.13) has a dual sum that is obtained with the use of the rule below.  We highlight this rule since it also applies to get certain groups of sums in Section 2.From [3], we recall the rule for the formation of the dual sum.(1)
Replace the even limit by the even limit corresponding to the other residue class modulo 4 and the odd limit by the odd limit corresponding to the other residue class modulo 4.(2)
Calculate the subscripts on the right in accordance with the paragraph following (2.13).(3)
Multiply the right side by-1.For example, for odd integerm, the dual of (2.13) is(2.16)∑n=4i4j+1(-1)(n+12)Umn+cWmn+d=-1Δ(VmV2m(2(j-i)+1)Xs+mV2m+2(-1)cXd-c),
where s is defined as before.Theorem 2.5.
Fix integersc,d, and  m.(i)
Ifc and d have the same parities, then
(2.17)∑n=4i+14j(-1)(n+12)W(2m+1)n+cW(2m+1)n+d=U2m+1U4(2m+1)(j-i)V2(2m+1)×(X2(2m+1)(j+i)+tX2(2m+1)(j+i)+t+1+pr(-1)t),∑n=4i+14j(-1)(n+12)W2mn+cW2mn+d=V2mU8m(j-i)V4m×(Wm(4j+4i)+tXm(4j+4i)+t),∑n=4i+34j(-1)(n+12)W(2m+1)n+cW(2m+1)n+d=U2m+1V2(2m+1)(2(j-i)-1)V2(2m+1)×(W(2m+1)(2(j+i)+1)+t+1X(2m+1)(2(j+i)+1)+t-r(-1)t),∑n=4i+34j(-1)(n+12)W2mn+cW2mn+d=V2mV4m(2(j-i)-1)ΔV4m×(Xm(4(j+i)+2)+t2+2r(-1)t)+2r(-1)cVd-cΔ,∑n=4i4j+3(-1)(n+12)W(2m+1)n+cW(2m+1)n+d=V2m+1U4(2m+1)(j-i+1)V2(2m+1)×(W(2m+1)(2(j+i)+1)+t+1X(2m+1)(2(j+i)+1)+t-r(-1)t),∑n=4i4j+3(-1)(n+12)W2mn+cW2mn+d=U2mU8m(j-i+1)V4m×(Xm(4(j+i)+2)+t2+2r(-1)t),∑n=4i+24j+3(-1)(n+12)W(2m+1)n+cW(2m+1)n+d=V2m+1V2(2m+1)(2(j-i)+1)ΔV2(2m+1)×(X2(2m+1)(j+i+1)+t+1X2(2m+1)(j+i+1)+t+pr(-1)t)-2r(-1)cVd-cΔ,∑n=4i+24j+3(-1)(n+12)W2mn+cW2mn+d=1V4m(U2mV4m(2(j-i)+1)W4m(j+i+1)+tX4m(j+i+1)+t),
where t=(c+d)/2+m.(ii)
Ifc and d have different parities, then
(2.18)∑n=4i+14j(-1)(n+12)W(2m+1)n+cW(2m+1)n+d=U2m+1U4(2m+1)(j-i)V2(2m+1)(X2(2m+1)(j+i)+v2+2r(-1)v),∑n=4i+14j(-1)(n+12)W2mn+cW2mn+d=V2mU8m(j-i)V4m(X4m(j+i)+v-1W4m(j+i)+v-r(-1)v),∑n=4i+34j(-1)(n+12)W(2m+1)n+cW(2m+1)n+d=1V2(2m+1)×U2m+1V2(2m+1)(2(j-i)-1)W(2m+1)(2(j+i)+1)+vX(2m+1)(2(j+i)+1)+v,∑n=4i+34j(-1)(n+12)W2mn+cW2mn+d=V2mV4m(2(j-i)-1)ΔV4m×(Xm(4(j+i)+2)+vXm(4(j+i)+2)+v-1+pr(-1)v)+2r(-1)cVd-cΔ,∑n=4i4j+3(-1)(n+12)W(2m+1)n+cW(2m+1)n+d=V2m+1U4(2m+1)(j-i+1)V2(2m+1)×X(2m+1)(2(j+i)+1)+vW(2m+1)(2(j+i)+1)+v,∑n=4i4j+3(-1)(n+12)W2mn+cW2mn+d=U2mU8m(j-i+1)V4m×(Xm(4(j+i)+2)+vXm(4(j+i)+2)+v-1-pr(-1)v),∑n=4i+24j+3(-1)(n+12)W(2m+1)n+cW(2m+1)n+d=-2r(-1)cVd-cΔ+V2m+1V2(2m+1)(2(j-i)+1)ΔV2(2m+1)(X2(2m+1)(j+i+1)+v2+2r(-1)v),∑n=4i+24j+3(-1)(n+12)W2mn+cW2mn+d=U2mV4m(2(j-i)+1)V4m×(W4m(j+i+1)+vX4m(j+i+1)+v-1-r(-1)v),
where r is defined as before and v=(c+d+1)/2+m.Proof.
Suppose thatc and d have the same parities. Consider
(2.19)∑n=4i+24j+3(-1)(n+12)W2mn+cW2mn+d=1Δ∑n=4i+24j+3(-1)(n+12)(A2α4mn+c+d+B2β4mn+c+d-ABα2mn+cβ2mn+d-ABβ2mn+cα2mn+d)=1Δ(b2∑n=4i+24j+3(-1)(n+12)V4mn+c+d+2ab∑n=4i+24j+3(-1)(n+12)V4mn+c+d-1+a2∑n=4i+24j+3(-1)(n+12)V4mn+c+d-2)+1Δ(-1)crVd-c∑n=4i+24j+3(-1)(n+12).
From the definition of {Vn}, we obtain
(2.20)∑n=4i+24j+3(-1)(n+12)V4mn+c=ΔU2mV4m(2(j-i)+1)U2m(4(j+i+1)+1)+cV4m.
Since
(2.21)∑n=4i+24j+3(-1)(n+12)=0,
we get
(2.22)∑n=4i+24j+3(-1)(n+12)W2mn+cW2mn+d=U2mV4m(2(j-i)+1)V4m(b2U2m(4(j+i+1)+1)+c+d+2abU2m(4(j+i+1)+1)+c+d-1+a2U2m(4(j+i+1)+1)+c+d-2)
Taking 2k=2m(4(j+i+1)+1)+c+d in Lemma 2.1, we write
(2.23)∑n=4i+24j+3(-1)(n+12)W2mn+cW2mn+d=U2mV4m(2(j-i)+1)Wm(4(j+i+1)+1)+(d+c)/2Xm(4(j+i+1)+1)+(d+c)/2ΔV4m.
Thus the result follows. Similar arguments yield the remaining formulas, where we must consider the parities of c,d.For example, the dual of (2.17) is given by if c and d have the same parities,(2.24)∑n=4i4j+1(-1)(n+12)W2mn+cW2mn+d=-U2mV4m(2(j-i)+1)W4m(j+i)+tX4m(j+i)+tV4m,
and the dual of (2.18) is given by if c and d have different parities, (2.25)∑n=4i4j+1(-1)(n+12)W2mn+cW2mn+d=-U2mV4m(2(j-i)+1)V4m×(W4m(j+i)+vX4m(j+i)+v-1-r(-1)v),
where t and v are defined as before.



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