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International Journal of Advances in Applied Mathematics and Mechanics

Generalized Tribonacci Numbers: Summing Formulas

Research Article

Yüksel Soykan

Department of Mathematics, Art and Science Faculty, Zonguldak Bülent Ecevit University, 67100, Zonguldak, Turkey

Received 04 February 2020; accepted (in revised version) 18 February 2020

Abstract: In this paper, closed forms of the summation formulas for generalized Tribonacci numbers are presented. As special cases, we give summation formulas of Tribonacci, Tribonacci-Lucas, Padovan, Perrin, Narayana and some other third order recurrance sequences.

MSC: 11B37 • 11B39 • 11B83

Keywords: Tribonacci numbers • Padovan numbers • Perrin numbers • Narayana numbers • Sum formulas

© 2020 The Author(s). This is an open access article under the CC BY-NC-ND license(https://creativecommons.org/licenses/by-nc-nd/3.0/).

1. Introduction

The generalized Tribonacci sequence {Wn(W0,W1,W2; r, s, t )}n≥0(or shortly {Wn}n≥0) (as a third-order generaliza- tion of Fibonacci numbers) is defined as follows:

Wn= r Wn−1+ sWn−2+ tWn−3, W0= a,W1= b,W2= c, n ≥ 3 (1)

where W0,W1,W2are arbitrary complex numbers and r, s, t are real numbers.

This sequence has been studied by many authors, see for example [1],[2],[3],[4],[5],[12],[14],[15],[16],[20],[29],[31],[32].

See also [7],[8],[30] for some work on second-order generalization of Fibonacci numbers.

The sequence {Wn}n≥0can be extended to negative subscripts by defining

W−n= −s

tW−(n−1)r

tW−(n−2)+1 tW−(n−3)

for n = 1,2,3,... when t 6= 0. Therefore, recurrence (1) holds for all integer n.

If we set r = s = t = 1 and W0= 0,W1= 1,W2= 1 then {Wn} is the well-known Tribonacci sequence and if we set r = s = t = 1 and W0= 3,W1= 1,W2= 3 then {Wn} is the well-known Tribonacci-Lucas sequence.

In fact, the generalized Tribonacci sequence is the generalization of the well-known sequences like Tribonacci, Tribonacci-Lucas, Padovan (Cordonnier), Perrin, Padovan-Perrin, Narayana, third order Jacobsthal and third order Jacobsthal-Lucas. In literature, for example, the following names and notations (seeTable 1) are used for the special case of r, s, t and initial values. Note that the sequence {Cn} is’t in the database of http://oeis.org [[17], yet.

In this work, we investigate summation formulas of generalized Tribonacci numbers. We present some works on summing formulas of the numbers in theTable 2.

E-mail address(es):[email protected]

57

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Table 1.A few special case of generalized Tribonacci sequences

Sequences (Numbers) Notation OEIS [17]

Tribonacci {Tn} = {Wn(0, 1, 1; 1, 1, 1)} A000073, A057597

Tribonacci-Lucas {Kn} = {Wn(3, 1, 3; 1, 1, 1)} A001644, A073145

third order Pell {Pn(3)} = {Wn(0, 1, 2; 2, 1, 1)} A077939, A077978

third order Pell-Lucas {Q(3)n } = {Wn(3, 2, 6; 2, 1, 1)} A276225, A276228

third order modified Pell {En(3)} = {Wn(0, 1, 1; 2, 1, 1)} A077997, A078049

Padovan (Cordonnier) {Pn} = {Wn(1, 1, 1; 0, 1, 1)} A000931

Perrin (Padovan-Lucas) {En} = {Wn(3, 0, 2; 0, 1, 1)} A001608, A078712

Padovan-Perrin {Sn} = {Wn(0, 0, 1; 0, 1, 1)} A000931, A176971

Pell-Padovan {Rn} = {Wn(1, 1, 1; 0, 2, 1)} A066983, A128587

Pell-Perrin {Cn} = {Wn(3, 0, 2; 0, 2, 1)} -

Jacobsthal-Padovan {Qn} = {Wn(1, 1, 1; 0, 1, 2)} A159284

Jacobsthal-Perrin (-Lucas) {Ln} = {Wn(3, 0, 2; 0, 1, 2)} A072328

Narayana {Nn} = {Wn(0, 1, 1; 1, 0, 1)} A078012

third order Jacobsthal {Jn(3)} = {Wn(0, 1, 1; 1, 1, 2)} A077947

third order Jacobsthal-Lucas { jn(3)} = {Wn(2, 1, 5; 1, 1, 2)} A226308

Table 2.A few special study of sum formulas

Name of sequence Papers which deal with summing formulas

Pell and Pell-Lucas [8],[10],[11]

Generalized Fibonacci [9],[22],[23],[21]

Generalized Tribonacci [6],[13],[18],[19]

Generalized Tetranacci [24],[25],[33]

Generalized Pentanacci [26],[27]

Generalized Hexanacci [28]

2. Sum formulas of Generalized Tribonacci Numbers with Positive Subscripts

The following theorem presents some summing formulas of generalized Tribonacci numbers with positive sub- scripts.

Theorem 2.1.

Let x be a complex number. For n ≥ 0, we have the following formulas:

(a) If t x3+ sx2+ r x − 1 6= 0, then

n

X

k=0

xkWk= Ω1

t x3+ sx2+ r x − 1 where

1 = xn+3Wn+3− (r x − 1) xn+2Wn+2− (sx2+ r x − 1)xn+1Wn+1

−x2W2+ x (r x − 1)W1+ (sx2+ r x − 1)W0. (b) If r2x − s2x2+ t2x3+ 2sx + 2r t x2− 1 6= 0 then

n

X

k=0

xkW2k= Ω2

r2x − s2x2+ t2x3+ 2sx + 2r t x2− 1 where

2 = −(sx − 1) xn+1W2n+2+ (t + r s) xn+2W2n+1+ t (r + t x) xn+2W2n +x (sx − 1)W2− (t + r s) x2W1+¡r2x − s2x2+ 2sx + r t x2− 1¢ W0. (c) If r2x − s2x2+ t2x3+ 2sx + 2r t x2− 1 6= 0 then

n

X

k=0

xkW2k+1= Ω3

r2x − s2x2+ t2x3+ 2sx + 2r t x2− 1 where

3 = (r + t x) xn+1W2n+2+¡s − s2x + t2x2+ r t x¢ xn+1W2n+1

−t (sx − 1) xn+1W2n− x (r + t x)W2+¡r2x + sx + r t x2− 1¢ W1+ t x (sx − 1)W0.

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Proof:

(a) Using the recurrence relation

Wn= r Wn−1+ sWn−2+ tWn−3

i.e.

tWn−3= Wn− r Wn−1− sWn−2

we obtain

t x0W0 = x0W3− r x0W2− sx0W1 t x1W1 = x1W4− r x1W3− sx1W2 t x2W2 = x2W5− r x2W4− sx2W3 t x3W3 = x3W6− r x3W5− sx3W4

...

t xn−3Wn−3 = xn−3Wn− r xn−3Wn−1− sxn−3Wn−2 t xn−2Wn−2 = xn−2Wn+1− r xn−2Wn− sxn−2Wn−1 t xn−1Wn−1 = xn−1Wn+2− r xn−1Wn+1− sxn−1Wn

t xnWn = xnWn+3− r xnWn+2− sxnWn+1. If we add the equations by side by, we get (a)

(b) and (c) Using the recurrence relation

Wn= r Wn−1+ sWn−2+ tWn−3

i.e.

r Wn−1= Wn− sWn−2− tWn−3

we obtain

r x1W3 = x1W4− sx1W2− t x1W1 r x2W5 = x2W6− sx2W4− t x2W3

r x3W7 = x3W8− sx3W6− t x3W5 r x4W9 = x4W10− sx4W8− t x4W7

...

r xn−1W2n−1 = xn−1W2n− sxn−1W2n−2− t xn−1W2n−3 r xnW2n+1 = xnW2n+2− sxnW2n− t xnW2n−1. Now, if we add the above equations by side by, we get

r (−x0W1+

n

X

k=0

xkW2k+1) = (xnW2n+2− x0W2− x−1W0+

n

X

k=0

xk−1W2k) (2)

−s(−x0W0+

n

X

k=0

xkW2k) − t(−xn+1W2n+1+

n

X

k=0

xk+1W2k+1).

Similarly, using the recurrence relation Wn= r Wn−1+ sWn−2+ tWn−3

i.e.

r Wn−1= Wn− sWn−2− tWn−3

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we write the following obvious equations;

r x1W2 = x1W3− sx1W1− t x1W0

r x2W4 = x2W5− sx2W3− t x2W2 r x3W6 = x3W7− sx3W5− t x3W4

r x4W8 = x4W9− sx4W7− t x4W6 r x5W10 = x5W11− sx5W9− t x5W8 r x6W12 = x6W13− sx6W11− t x6W10

...

r xn−1W2n−2 = xn−1W2n−1− sxn−1W2n−3− t xn−1W2n−4 r xnW2n = xnW2n+1− sxnW2n−1− t xnW2n−2 Now, if we add the above equations by side by, we obtain

r (−W0+

n

X

k=0

xkW2k) = (−W1+

n

X

k=0

xkW2k+1) − s(−xn+1W2n+1+

n

X

k=0

xk+1W2k+1) (3)

−t (−xn+1W2n+

n

X

k=0

xk+1W2k).

Then, solving the system (2)-(3), the required result of (b) and (c) follow.

2.1. The case x = 1

The case x = 1 ofTheorem 2.1is given in [18], see also [19]. In this subsection, we only consider the case x = 1,r = 0, s = 2, t = 1 and we present a theorem which is in different forms than given in [18]).

Observe that setting x = 1,r = 0, s = 2, t = 1 (i.e. for the generalized Pell-Padovan case) inTheorem 2.1(b) and (c) makes the right hand side of the sum formulas to be an indeterminate form. Application of L’Hospital rule however provides the evaluation of the sum formulas. If r = 0, s = 2, t = 1 then we have the following theorem.

Theorem 2.2.

If r = 0, s = 2, t = 1 then for n ≥ 0 we have the following formulas:

(a) Pn

k=0Wk=12(Wn+3+ Wn+2− Wn+1− W2− W1+ W0) . (b) Pn

k=0W2k= (n + 3)W2n+2− (n + 2)W2n+1− (n + 3)W2n− 3W2+ 2W1+ 4W0. (c) Pn

k=0W2k+1= −(n + 2)W2n+2+ (n + 3)W2n+1+ (n + 3)W2n+ 2W2− 2W1− 3W0. Proof:

(a) Taking r = 0, s = 2, t = 1 inTheorem 2.1(a) we obtain (a).

(b) We useTheorem 2.1(b). If we set r = 0, s = 2, t = 1 inTheorem 2.1(b) then we have Pn

k=0xkW2k=(2x−1)xn+1W2n+2−xn+2W2n+1−xn+3W2n−x(2x−1)W2+x

2W1+(4x2−4x+1)W0

−x3+4x2−4x+1 .

For x = 1, the right hand side of the above sum formulas is an indeterminate form. Now, we can use L’Hospital rule. Then we get

n

X

k=0

W2k =

d d x( f1(x))

d

d x(−x3+ 4x2− 4x + 1)

¯

¯

¯

¯

¯x=1

= (n + 3)W2n+2− (n + 2)W2n+1− (n + 3)W2n− 3W2+ 2W1+ 4W0

where

f1(x) = (2x − 1) xn+1W2n+2− xn+2W2n+1− xn+3W2n− x (2x − 1)W2+ x2W1+¡4x2− 4x + 1¢ W0.

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(c) We useTheorem 2.1(c). If we set r = 0, s = 2, t = 1 inTheorem 2.1(c) then we have Pn

k=0xkW2k+1=−xn+2W2n+2(x2−4x+2)xn+1W2n+1+(2x−1)xn+1W2n+x2W2−(2x−1)W1−x(2x−1)W0

−x3+4x2−4x+1 .

For x = 1, the right hand side of the above sum formulas is an indeterminate form. Now, we can use L’Hospital rule. Then we get

n

X

k=0

W2k+1 =

d d x( f2(x))

d

d x(−x3+ 4x2− 4x + 1)

¯

¯

¯

¯

¯x=1

= −(n + 2)W2n+2+ (n + 3)W2n+1+ (n + 3)W2n+ 2W2− 2W1− 3W0

where

f2(x) = −xn+2W2n+2¡x2− 4x + 2¢ xn+1W2n+1+ (2x − 1) xn+1W2n+ x2W2− (2x − 1)W1− x (2x − 1)W0

From the above theorem we have the following corollary which gives sum formulas of Pell-Padovan numbers (take Wn= Rnwith R0= 1, R1= 1, R2= 1).

Corollary 2.1.

For n ≥ 0, Pell-Padovan numbers have the following property:

(a) Pn

k=0Rk=12(Rn+3+ Rn+2− Rn+1− 1) . (b) Pn

k=0R2k= (n + 3) R2n+2− (n + 2) R2n+1− (n + 3) R2n+ 3.

(c) Pn

k=0R2k+1= −(n + 2) R2n+2+ (n + 3) R2n+1+ (n + 3) R2n− 3.

Taking Wn= Cnwith C0= 3,C1= 0,C2= 2 in the last theorem, we have the following corollary which presents sum formulas of Pell-Perrin numbers.

Corollary 2.2.

For n ≥ 0, Pell-Perrin numbers have the following property:

(a) Pn

k=0Ck=12(Cn+3+Cn+2−Cn+1+ 1) . (b) Pn

k=0C2k= (n + 3)C2n+2− (n + 2)C2n+1− (n + 3)C2n+ 6.

(c) Pn

k=0C2k+1= −(n + 2)C2n+2+ (n + 3)C2n+1+ (n + 3)C2n− 5.

2.2. The case x = −1

We now consider the case x = −1 inTheorem 2.1.

Taking x = −1,r = s = t = 1 inTheorem 2.1(a), (b) and (c), we obtain the following proposition.

Proposition 2.1.

If r = s = t = 1 then for n ≥ 0 we have the following formulas:

(a) Pn

k=0(−1)kWk=12((−1)n(Wn+3− 2Wn+2+ Wn+1) + W2− 2W1+ W0).

(b) Pn

k=0(−1)kW2k=12((−1)nW2n+2− (−1)nW2n+1− W2+ W1+ 2W0).

(c) Pn

k=0(−1)kW2k+1=12((−1)nW2n+1+ (−1)nW2n+ W1− W0).

From the above proposition, we have the following corollary which gives sum formulas of Tribonacci numbers (take Wn= Tnwith T0= 0, T1= 1, T2= 1).

Corollary 2.3.

For n ≥ 0, Tribonacci numbers have the following properties.

(a) Pn

k=0(−1)kTk=12((−1)n(Tn+3− 2Tn+2+ Tn+1) − 1).

(b) Pn

k=0(−1)kT2k=12(−1)n(T2n+2− T2n+1).

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(c) Pn

k=0(−1)kT2k+1=12((−1)nT2n+1+ (−1)nT2n+ 1).

Taking Wn= Knwith K0= 3, K1= 1, K2= 3 in the above proposition, we have the following corollary which presents sum formulas of Tribonacci-Lucas numbers.

Corollary 2.4.

For n ≥ 0, Tribonacci-Lucas numbers have the following properties.

(a) Pn

k=0(−1)kKk=12((−1)n(Kn+3− 2Kn+2+ Kn+1) + 4).

(b) Pn

k=0(−1)kK2k=12((−1)nK2n+2− (−1)nK2n+1+ 4).

(c) Pn

k=0(−1)kK2k+1=12((−1)nK2n+1+ (−1)nK2n− 2).

Taking x = −1,r = 2, s = 1, t = 1 inTheorem 2.1(a), (b) and (c), we obtain the following proposition.

Proposition 2.2.

If r = 2, s = 1, t = 1 then for n ≥ 0 we have the following formulas:

(a) Pn

k=0(−1)kWk=13((−1)n(Wn+3− 3Wn+2+ 2Wn+1) + W2− 3W1+ 2W0).

(b) Pn

k=0(−1)kW2k=15((−1)n(2W2n+2− 3W2n+1− W2n) − 2W2+ 3W1+ 6W0).

(c) Pn

k=0(−1)kW2k+1=15((−1)n(W2n+2+ W2n+1+ 2W2n) − W2+ 4W1− 2W0).

From the last proposition, we have the following corollary which gives sum formulas of third-order Pell numbers (take Wn= Pnwith P0= 0, P1= 1, P2= 2).

Corollary 2.5.

For n ≥ 0, third-order Pell numbers have the following properties:

(a) Pn

k=0(−1)kPk=13((−1)n(Pn+3− 3Pn+2+ 2Pn+1) − 1).

(b) Pn

k=0(−1)kP2k=15((−1)n(2P2n+2− 3P2n+1− P2n) − 1).

(c) Pn

k=0(−1)kP2k+1=15((−1)n(P2n+2+ P2n+1+ 2P2n) + 2).

Taking Wn= Qnwith Q0= 3,Q1= 2,Q2= 6 in the last proposition, we have the following corollary which presents sum formulas of third-order Pell-Lucas numbers.

Corollary 2.6.

For n ≥ 0, third-order Pell-Lucas numbers have the following properties:

(a) Pn

k=0(−1)kQk=13((−1)n(Qn+3− 3Qn+2+ 2Qn+1) + 6).

(b) Pn

k=0(−1)kQ2k=15((−1)n(2Q2n+2− 3Q2n+1−Q2n) + 12).

(c) Pn

k=0(−1)kQ2k+1=15((−1)n(Q2n+2+Q2n+1+ 2Q2n) − 4).

From the last proposition, we have the following corollary which gives sum formulas of third-order modified Pell numbers (take Wn= Enwith E0= 0, E1= 1, E2= 1).

Corollary 2.7.

For n ≥ 0, third-order modified Pell numbers have the following properties:

(a) Pn

k=0(−1)kEk=13((−1)n(En+3− 3En+2+ 2En+1) − 2).

(b) Pn

k=0(−1)kE2k=15((−1)n(2E2n+2− 3E2n+1− E2n) + 1).

(c) Pn

k=0(−1)kE2k+1=15((−1)n(E2n+2+ E2n+1+ 2E2n) + 3).

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Taking x = −1,r = 0, s = 1, t = 1 inTheorem 2.1(a), (b) and (c), we obtain the following proposition.

Proposition 2.3.

If r = 0, s = 1, t = 1 then for n ≥ 0 we have the following formulas:

(a) Pn

k=0(−1)kWk= (−1)n(Wn+3− Wn+2) + W2− W1. (b) Pn

k=0(−1)kW2k=15((−1)n(2W2n+2− W2n+1+ W2n) − 2W2+ W1+ 4W0).

(c) Pn

k=0(−1)kW2k+1=15((−1)n(−W2n+2+ 3W2n+1+ 2W2n) + W2+ 2W1− 2W0).

From the last proposition, we have the following corollary which gives sum formulas of Padovan numbers (take Wn= Pnwith P0= 1, P = 1, P2= 1).

Corollary 2.8.

For n ≥ 0, Padovan numbers have the following properties.

(a) Pn

k=0(−1)kPk= (−1)n(Pn+3− Pn+2).

(b) Pn

k=0(−1)kP2k=15((−1)n(2P2n+2− P2n+1+ P2n) + 3).

(c) Pn

k=0(−1)kP2k+1=15((−1)n(−P2n+2+ 3P2n+1+ 2P2n) + 1).

Taking Wn= En with E0= 3, E2= 0, E2= 2 in the last proposition, we have the following corollary which presents sum formulas of Perrin numbers.

Corollary 2.9.

For n ≥ 0, Perrin numbers have the following properties.

(a) Pn

k=0(−1)kEk= (−1)n(En+3− En+2) + 2.

(b) Pn

k=0(−1)kE2k=15((−1)n(2E2n+2− E2n+1+ E2n) + 8).

(c) Pn

k=0(−1)kE2k+1=15((−1)n(−E2n+2+ 3E2n+1+ 2E2n) − 4).

Taking Wn= Snwith S0= 0, S2= 0, S2= 1 in the last proposition, we have the following corollary which gives sum formulas of Padovan-Perrin numbers.

Corollary 2.10.

For n ≥ 0, Padovan-Perrin numbers have the following properties.

(a) Pn

k=0(−1)kSk= (−1)n(Sn+3− Sn+2) + 1.

(b) Pn

k=0(−1)kS2k=15((−1)n(2S2n+2− S2n+1+ S2n) − 2).

(c) Pn

k=0(−1)kS2k+1=15((−1)n(−S2n+2+ 3S2n+1+ 2S2n) + 1).

If x = −1,r = 0, s = 2, t = 1 then t x3+ sx2+ r x − 1 = 0 so we can’t useTheorem 2.1(a) directly. Observe that setting x = −1,r = 0, s = 2, t = 1 (i.e. for the generalized Pell-Padovan case) inTheorem 2.1(a) makes the right hand side of the sum formulas to be an indeterminate form. Application of L’Hospital rule provides the evaluation of the sum formulas. If r = 0, s = 2, t = 1 then we have the following theorem.

Theorem 2.3.

If r = 0, s = 2, t = 1 then for n ≥ 0 we have the following formulas:

(a) Pn

k=0(−1)kWk= (−1)n(−(n + 3)Wn+3+ (n + 2)Wn+2+ (n + 5)Wn+1) − 2W2+ W1+ 4W0. (b) Pn

k=0(−1)kW2k=101(3 (−1)nW2n+2− (−1)nW2n+1+ (−1)nW2n− 3W2+ W1+ 9W0).

(c) Pn

k=0(−1)kW2k+1=101((−1)n(−W2n+2+ 7W2n+1+ 3W2n) + 3W1− 3W0+ W2).

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Proof:

(a) We useTheorem 2.1(a). If we set r = 0, s = 2, t = 1 inTheorem 2.1(a) then we have

n

X

k=0

xkWk=xn+3Wn+3+ xn+2Wn+2− xn+1¡2x2− 1¢ Wn+1− x2W2− xW1+¡2x2− 1¢ W0

x3+ 2x2− 1 .

For x = −1, the right hand side of the above sum formulas is an indeterminate form. Now, we can use L’Hospital rule. Then we get

n

X

k=0

(−1)kWk =

d

d x(xn+3Wn+3+ xn+2Wn+2− xn+1¡2x2− 1¢ Wn+1− x2W2− xW1+¡2x2− 1¢ W0)

d

d x(x3+ 2x2− 1)

¯

¯

¯

¯

¯x=−1

= (−1)n(−(n + 3)Wn+3+ (n + 2)Wn+2+ (n + 5)Wn+1) − 2W2+ W1+ 4W0.

(b) Taking r = 0, s = 2, t = 1 inTheorem 2.1(b) we obtain (b).

(c) Taking r = 0, s = 2, t = 1 inTheorem 2.1(c) we obtain (c).

From the above theorem we have the following corollary which gives sum formulas of Pell-Padovan numbers (take Wn= Rnwith R0= 1, R1= 1, R2= 1).

Corollary 2.11.

For n ≥ 0, Pell-Padovan numbers have the following property:

(a) Pn

k=0(−1)kRk= (−1)n(−(n + 3)Rn+3+ (n + 2) Rn+2+ (n + 5) Rn+1) + 3.

(b) Pn

k=0(−1)kR2k=101(3 (−1)nR2n+2− (−1)nR2n+1+ (−1)nR2n+ 7).

(c) Pn

k=0(−1)kR2k+1=101((−1)n(−R2n+2+ 7R2n+1+ 3R2n) + 1).

Taking Wn= Cnwith C0= 3,C1= 0,C2= 2 in the last theorem, we have the following corollary which presents sum formulas of Pell-Perrin numbers.

Corollary 2.12.

For n ≥ 0, Pell-Perrin numbers have the following property:

(a) Pn

k=0(−1)kCk= (−1)n(−(n + 3)Cn+3+ (n + 2)Cn+2+ (n + 5)Cn+1) + 8.

(b) Pn

k=0(−1)kC2k=101(3 (−1)nC2n+2− (−1)nC2n+1+ (−1)nC2n+ 21).

(c) Pn

k=0(−1)kC2k+1=101((−1)n(−C2n+2+ 7C2n+1+ 3C2n) − 7).

Taking x = −1,r = 0, s = 1, t = 2 inTheorem 2.1(a), (b) and (c), we obtain the following proposition.

Proposition 2.4.

If r = 0, s = 1, t = 2 then for n ≥ 0 we have the following formulas:

(a) Pn

k=0(−1)kWk=12((−1)nWn+3− (−1)nWn+2+ W2− W1).

(b) Pn

k=0(−1)kW2k=14((−1)n(W2n+2− W2n+1+ 2W2n) − W2+ W1+ 2W0).

(c) Pn

k=0(−1)kW2k+1=14((−1)n(−W2n+2+ 3W2n+1+ 2W2n) + W2+ W1− 2W0)

Taking Wn= Qnwith Q0= 1,Q1= 1,Q2= 1 in the last proposition, we have the following corollary which presents sum formulas of Jacobsthal-Padovan numbers.

Corollary 2.13.

For n ≥ 0, Jacobsthal-Padovan numbers have the following properties.

(a) Pn

k=0(−1)kQk=12(−1)n(Qn+3−Qn+2).

(9)

(b) Pn

k=0(−1)kQ2k=14((−1)n(Q2n+2−Q2n+1+ 2Q2n) + 2).

(c) Pn

k=0(−1)kQ2k+1=14(−1)n(−Q2n+2+ 3Q2n+1+ 2Q2n).

From the last proposition, we have the following corollary which gives sum formulas of Jacobsthal-Perrin numbers (take Wn= Lnwith L0= 3, L1= 0, L2= 2).

Corollary 2.14.

For n ≥ 0, Jacobsthal-Perrin numbers have the following properties.

(a) Pn

k=0(−1)kLk=12((−1)nLn+3− (−1)nLn+2+ 2).

(b) Pn

k=0(−1)kL2k=14((−1)n(L2n+2− L2n+1+ 2L2n) + 4).

(c) Pn

k=0(−1)kL2k+1=14((−1)n(−L2n+2+ 3L2n+1+ 2L2n) − 4).

Taking x = −1,r = 1, s = 0, t = 1 inTheorem 2.1(a), (b) and (c), we obtain the following proposition.

Proposition 2.5.

If r = 1, s = 0, t = 1 then for n ≥ 0 we have the following formulas:

(a) Pn

k=0(−1)kWk=13((−1)n(Wn+3− 2Wn+2+ 2Wn+1) + W2− 2W1+ 2W0).

(b) Pn

k=0(−1)kW2k= (−1)n(W2n+2− W2n+1) − W2+ W1+ W0. (c) Pn

k=0(−1)kW2k+1= (−1)nW2n+ W1− W0.

From the last proposition, we have the following corollary which presents sum formulas of Narayana numbers (take Wn= Nnwith N0= 0, N1= 1, N2= 1).

Corollary 2.15.

For n ≥ 0, Narayana numbers have the following properties.

(a) Pn

k=0(−1)kNk=13((−1)n(Nn+3− 2Nn+2+ 2Nn+1) − 1).

(b) Pn

k=0(−1)kN2k= (−1)n(N2n+2− N2n+1).

(c) Pn

k=0(−1)kN2k+1= (−1)nN2n+ 1.

Taking x = −1,r = 1, s = 1, t = 2 inTheorem 2.1(a), (b) and (c), we obtain the following proposition.

Proposition 2.6.

If r = 1, s = 1, t = 2 then for n ≥ 0 we have the following formulas:

(a) Pn

k=0(−1)kWk=13((−1)n(Wn+3− 2Wn+2+ Wn+1) + W2− 2W1+ W0).

(b) Pn

k=0(−1)kW2k=15((−1)n(2W2n+2− 3W2n+1+ 2W2n) − 2W2+ 3W1+ 3W0).

(c) Pn

k=0(−1)kW2k+1=15((−1)n(−W2n+2+ 4W2n+1+ 4W2n) + W2+ W1− 4W0).

Taking Wn= Jnwith J0= 0, J1= 1, J2= 1 in the last proposition, we have the following corollary which presents sum formulas of third order Jacobsthal numbers.

Corollary 2.16.

For n ≥ 0, third order Jacobsthal numbers have the following properties.

(a) Pn

k=0(−1)kJk=13((−1)n(Jn+3− 2Jn+2+ Jn+1) − 1).

(b) Pn

k=0(−1)kJ2k=15((−1)n(2J2n+2− 3J2n+1+ 2J2n) + 1).

(c) Pn

k=0(−1)kJ2k+1=15((−1)n(−J2n+2+ 4J2n+1+ 4J2n) + 2).

(10)

From the last proposition, we have the following corollary which gives sum formulas of third order Jacobsthal-Lucas numbers (take Wn= jnwith j0= 2, j1= 1, j2= 5).

Corollary 2.17.

For n ≥ 0, third order Jacobsthal-Lucas numbers have the following properties.

(a) Pn

k=0(−1)kjk=13((−1)n( jn+3− 2 jn+2+ jn+1) + 5).

(b) Pn

k=0(−1)kj2k=15((−1)n(2 j2n+2− 3 j2n+1+ 2 j2n) − 1).

(c) Pn

k=0(−1)kj2k+1=15((−1)n(−j2n+2+ 4 j2n+1+ 4 j2n) − 2).

2.3. The Case x = 1 + i

We now consider the complex case x = 1 + i inTheorem 2.1.

Taking x = 1 + i ,r = s = t = 1 inTheorem 2.1(a), (b) and (c), we obtain the following proposition.

Proposition 2.7.

If r = s = t = 1 then for n ≥ 0 we have the following formulas:

(a) Pn

k=0(1 + i )kWk=−2+5i1 ((1 + i )n(3 (1 − i )Wn+1+ 2Wn+2+ 2 (−1 + i )Wn+3) − 2iW2− (1 − i )W1+ 3iW0).

(b) Pn

k=0(1 + i )kW2k=7i1((1 + i )n((1 − i )W2n+2+ 4iW2n+1+ 2(−1 + 2i )W2n) − (1 − i )W2− 4iW1+ (2 + 3i )W0) (c) Pn

k=0(1 + i )kW2k+1=7i1((1 + i )n((1 + 3i )W2n+2+ (−1 + 3i )W2n+1+ (1 − i )W2n) − (1 + 3i )W2+ (1 + 4i )W1− (1 − i )W0).

From the above proposition, we have the following corollary which gives sum formulas of Tribonacci numbers (take Wn= Tnwith T0= 0, T1= 1, T2= 1).

Corollary 2.18.

For n ≥ 0, Tribonacci numbers have the following properties.

(a) Pn

k=0(1 + i )kTk=−2+5i1 ((1 + i )n(3 (1 − i )Tn+1+ 2Tn+2+ 2 (−1 + i ) Tn+3) − 1 − i ).

(b) Pn

k=0(1 + i )kT2k=7i1((1 + i )n((1 − i )T2n+2+ 4i T2n+1+ 2(−1 + 2i )T2n) − 1 − 3i ).

(c) Pn

k=0(1 + i )kT2k+1=7i1((1 + i )n((1 + 3i )T2n+2+ (−1 + 3i )T2n+1+ (1 − i )T2n) + i ).

Taking Wn= Knwith K0= 3, K1= 1, K2= 3 in the last proposition, we have the following corollary which presents sum formulas of Tribonacci-Lucas numbers.

Corollary 2.19.

For n ≥ 0, Tribonacci-Lucas numbers have the following properties.

(a) Pn

k=0(1 + i )kKk=−2+5i1 ((1 + i )n(3 (1 − i )Kn+1+ 2Kn+2+ 2 (−1 + i ) Kn+3) − 1 + 4i ).

(b) Pn

k=0(1 + i )kK2k=7i1((1 + i )n((1 − i )K2n+2+ 4i K2n+1+ 2(−1 + 2i )K2n) + 3 + 8i ).

(c) Pn

k=0(1 + i )kK2k+1=7i1((1 + i )n((1 + 3i )K2n+2+ (−1 + 3i )K2n+1+ (1 − i )K2n) − 5 − 2i ).

Corresponding sums of the other third order linear sequences can be calculated similarly when x = 1 + i .

References

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