On Sums of Products of Horadam Numbers

Transcription

1 KYUNGPOOK Math J 49(009), On Sums of Products of Horadam Numbers Zvonko ƒerin Kopernikova 7, Zagreb, Croatia, Europe cerin@mathhr Abstract In this paper we give formulae for sums of products of two Horadam type generalized Fibonacci numbers with the same recurrence equation and with possibly different initial conditions Analogous improved alternating sums are also studied as well as various derived sums when terms are multiplied either by binomial coecients or by members of the sequence of natural numbers These formulae are related to the recent work of Belbachir and Bencherif, ƒerin and ƒerin and Gianella 1 Introduction The generalized Fibonacci sequence {w n } = {w n (a 0, b 0 ; p, q)} is dened by w 0 = a 0, w 1 = b 0, w n = p w n 1 q w n (n ), where a 0, b 0, p and q are arbitrary complex numbers, with q 0 The numbers w n have been studied by Horadam (see, eg [10]) A useful and interesting special cases are {U n } = {w n (0, 1; p, q)} and {V n } = {w n (, p; p, q)} that were investigated by Lucas [11] For integers a 0, c 0, 0, b > 0 and d > 0, let P = U a+b U c+d, Q = U a+b V c+d and R = V a+b V c+d In [1] some formulae for the sums n =0 P, n =0 Q, n =0 R, n =0 ( 1) P, n =0 ( 1) Q and n =0 ( 1) R have been discovered in the special case when b = d = and q = ±1 Even in these restricted case they gave unication of earlier results by ƒerin and by ƒerin and Gianella for Fibonacci, Lucas, Pell and Pell-Lucas numbers (see [3] [9]) In [] the author eliminated all restrictions from the article [1] on b, d and q (except that q 0) Some other types of sums have also been studied like the improved alternating sums (when we multiply terms by increasing powers of a xed complex number), the sums with binomial coecients and sums in which we multiply terms by increasing natural numbers The goal in this paper is to extend these results to Horadam type generalized Fibonacci numbers Even in this more general case these sums could be evaluated using the sum of a geometric series Received May 6, 008; accepted May 6, Mathematics Subect Classication: Primary 11B39, 11Y55, 05A19 Key words and phrases: Horadam generalized Fibonacci numbers, Lucas numbers, sums of products 483

2 484 Zvonko čerin Sums of products of two Horadam numbers We rst want to nd the formula for the sum Ψ 1 = w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) =0 when a 0, b 0, c 0, d 0, p and q 0 are complex numbers and n 0, a 0, c 0, b > 0 and d > 0 are integers Let α and β be the roots of x p x + q = 0 Then α = p + and β = p, where = p 4 q Moreover, α β =, α + β = p, α β = q and the Binet forms of w n, U n and V n are w n = (b 0 a 0 β) α n + (a 0 α b 0 ) β n, U n = αn β n α β α β, V n = α n + β n, if α β, and w n = α n 1 (a 0 α + n (b 0 a 0 α)), U n = n α n 1, V n = α n, if α = β Let A 1 = b 0 a 0 α, A = d 0 c 0 α, B 1 = a 0 β b 0, B = c 0 β d 0 Let E = α b+d, F = α b β d, G = α d β b and H = β b+d Let e = α a+c B 1 B, f = α a β c A B 1, g = α c β a A 1 B and h = β a+c A 1 A When E 1, for any integer n 0, let E n = En+1 1 E 1 We similarly dene F n, G n and H n On the other hand, when α b β b, for any integer n 0, let b n = αb(n+1) β b(n+1) and α b n (α b β b b n = αb(n+1) β b(n+1) ) β b n (α b β b ) We similarly dene d n and d n For any integer n 0, let λ n = n + 1 Let T = α a+c Let C 1 = A 1 a + a 0 α, C = A c + c 0 α, K 1 = b d A 1 A, K 3 = C 1 C and K = b A 1 C + d A C 1 Let K 4 = K 1 + K + K 3 Let K, M, N and P be n ( n + 1) K n K + 6 K 3, n E n+3 ( n + n 1) E n+ + (n + 1) E n+1 E(E + 1), n E n+3 ( n + 1) E n+ + (n + 1) E n+1 + E(E 1), E n+3 E n+ + E n+1 (E 1) (n + 1) K T Theorem 1 (a) When = 0 and E = 1, then Ψ 1 = 6 (b) When = 0 and E 1, then Ψ 1 = T [ K 1 M + K N + K 3 P ] (E 1) 3 Proof (a) Recall that when = 0, then w a+b (a 0, b 0 ; p, q) = α a 1+b [ b A 1 + C 1 ]

3 On Sums of Products of Horadam Numbers 485 and w c+d (c 0, b 0 ; p, q) = α c 1+d [ d A + C ] Since, E = α b+d = 1, we see that the product is equal to =0 1 = n + 1, n =0 follows that Ψ 1 has the above value (b) Since = 0, the product is equal to =0 E = w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) T [ K 1 + K + K 3 ] n(n + 1) =, and n =0 = w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) T E [ K 1 + K + K 3 ] P (E 1), n 3 =0 E = it follows that Ψ 1 has the above value n(n + 1)( n + 1), it 6 N (E 1) 3, and n =0 E = M (E 1) 3, The following theorem covers for the sum Ψ 1 the cases when 0 It uses Table 1 that should be read as follows The symbols and in column E mean E 1 and E = 1 In column b they mean α b β b and α b = β b In columns F, G, H and d they have analogous meanings The third subcase should be read as follows: When ( 0), E = 1 and α b = β b, then G = 1 and H = F and for F 1 the product Ψ 1 is equal to λ n (e + g) + F n (f + h) Theorem When 0, then Table 1 gives the value of Ψ 1 In all other cases the product Ψ 1 is equal to λ n (e + f + g + h) Proof of row 1 When 0, we have w a+b (a 0, b 0 ; p, q) = 1 [ α a B 1 (α b ) + β a A 1 (β b ) ] and w c+d (c 0, d 0 ; p, q) = 1 [ α c B (α d ) + β c A (β d ) ] Hence, the product w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) is equal to e E + f F + g G + h H =0 E = E n, we get Ψ 1 = e E n + f F n + g G n + h H n

4 486 Zvonko čerin E F G H b d Ψ 1 1 E n e + F n f + G n g + H n h λ n e + F n f + b n g + H n h 3 F λ n (e + g) + F n (f + h) 4 λ n e + d n f + G n g + H n h 5 G λ n (e + f) + G n (g + h) 6 (see 5) 7 (see 3) 8 λ n (e + g) + d n (f + h) 9 λ n (e + h) + d n f + d n g 10 E n e + λ n f + G n g + b n h 11 E E n (e + g) + λ n (f + h) 1 d n e + λ n f + G n g + H n h 13 G λ n (e + f) + G n (g + h) 14 d n e + λ n (f + g) + d n h 15 (see 11) 16 d n (e + g) + λ n (f + h) 17 b n e + F n f + λ n g + H n h 18 F λ n (e + g) + F n (f + h) 19 E n e + F n f + λ n g + d n h 0 E E n (e + f) + λ n (g + h) 1 b n (e + f) + λ n (g + h) E n e + b n f + G n g + λ n h 3 E E n (e + g) + λ n (f + h) 4 E n e + F n f + d n g + λ n h 5 E E n (e + f) + λ n (g + h) Table 1: The product Ψ 1 when 0

5 On Sums of Products of Horadam Numbers 487 Proof of row When 0 and E = α b+d = 1, we get w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) = e + ff + g ( β b α b ) + hh =0 1 = λ n, n =0 F = F n and n =0 that Ψ 1 = e λ n + f F n + g b n + h H n Proof of row 3 When 0, E = α b+d = 1 and α b = β b, then and H = β b β d = α b β d = F Hence, ( ) β b = b n (for α b β b ), it follows α b G = β b α d = α b α d = E = 1 w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) = e + g + (f + h)f =0 1 = λ n and n =0 F = F n (for F 1, of course), it follows that the product Ψ 1 is equal to (e + g) λ n + (f + h) F n The missing case in the Table 1 after the third row is clearly when E = 1, α b = β b and F = 1 The above product is w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) = e + f + g + h, so that Ψ 1 = λ n (e + f + g + h) The selection p = 0, q = 1, b = and d = shows that this case can actually happen Ũ n+1 Notice that α n = V n + U n and β n = V n U n for 0 and α n = β n = n + 1 = Ṽn for = 0 Hence, it is clear that each of the above expressions for the sum Ψ 1 could be transformed into an expression in Lucas numbers U n and V n (or Ũ n and Ṽn) In most cases these formulae are more complicated then the ones given above This applies also to other sums that we consider in this paper 3 Sum with binomial coecients In this section we consider the sum ( ) n Ψ = w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q), =0 when a 0, b 0, c 0, d 0, p and q 0 are complex numbers and n 0, a 0, c 0, b > 0 and d > 0 are integers Let V and U be n E [(n E + 1) K 1 + (E + 1) K ] + (E + 1) K 3 and E K 4 + K 3

6 488 Zvonko čerin Theorem 3 (a) When = 0, then (b) When 0, then Proof (b) Since T K 3, if n = 0, Ψ = T U, if n = 1, T (E + 1) n V, if n, Ψ = (E + 1)n e + (F + 1) n f + (G + 1) n g + (H + 1) n h ( ) n w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) = ( ) n e E + f F + g G + h H, from n ( n ) =0 E = (E + 1) n, it follows that Ψ indeed has the above value 4 The improved alternating sums, I In this section we consider the sums obtained from the sums Ψ 1 and Ψ by multiplication of their terms with the powers of a xed complex number k When k = 1 we obtain the familiar alternating sums More precisely, we study the sums Ψ 3 = Ψ 4 = =0 k w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q), =0 ( ) n k w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q), when a 0, b 0, c 0, d 0, p and q 0 are complex numbers and n 0, a 0, c 0, b > 0 and d > 0 are integers Let E = k α b+d, F = k α b β d, G = k α d β b and H = k β b+d When E 1, for any integer n 0, let E n = En+1 1 E 1 We similarly dene F n, G n and H n In this section we can assume that k 1 and k 0 because the case when k = 1 was treated earlier while for k = 0 all sums are equal to zero With this new meaning of the symbols E, F, G and H we have the following result Theorem 4 (a) The values given in Theorems 1 and express the sum Ψ 3 In particular, when 0, then the Table 1 gives the values of Ψ 3 In all other cases the product Ψ 3 is equal to λ n (e + f + g + h) (b) The values given in Theorem 3 for the sums Ψ express also the sum Ψ 4

7 On Sums of Products of Horadam Numbers 489 Proof (b) Since ( ) ( ) n n e E k + f F + g G + h H w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) =, from n ( n ) =0 E = (E + 1) n, it follows that Ψ 4 indeed has the same expression as the sum Ψ 5 Terms multiplied by natural numbers In this section we study the sums Ψ 5 = ( + 1) w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q), Ψ 6 = =0 =0 ( ) n ( + 1) w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q), when a 0, b 0, c 0, d 0, p and q 0 are complex numbers and n 0, a 0, c 0, b > 0 and d > 0 are integers Let E = α b+d, F = α b β d, G = α d β b, H = β b+d Let e = α a+c B 1 B, f = α a β c A B 1, g = α c β a A 1 B, h = β a+c A 1 A When E 1, for any integer n 0, let E n = (n + 1)En+ (n + )E n (E 1) We similarly dene F n, G n and H n On the other hand, when α b β b, for any integer n 0, let and b n = αb(n+) + (n + 1)β b(n+) (n + )α b β n+1 α b n (α b β b ) b n = βb(n+) + (n + 1)α b(n+) (n + )β b α n+1 β b n (α b β b ) We similarly dene d n and d (n + 1)(n + ) n For any integer n 0, let λ n = Let M and N denote n E [ n+3 n (n + 1) E ( 3 n + 6 n 1 )] + (n + ) E n+1 [ (3 n + 3 n ) E (n + 1) ] + E ( E + 1) and (E 1) [ n (n + 1) E n+3 n (n + ) E n+ +(n + ) (n + 1) E n+1 E ] Theorem 5 (a) When = 0 and E = 1, then the sum Ψ 5 is equal to λ n T [ n(3 n + 1) K n K + 6 K 3 ] 6 (b) When = 0 and E 1, then the sum Ψ 5 is equal to T (E 1) 4 [ K1 M + K N + K 3 (E 1) 4 E n ]

8 490 Zvonko čerin Proof (b) Since = 0, we have ( + 1) w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) = ( + 1) ( α a+b 1 [ b A 1 + C 1 ] ) ( α c+d 1 [ d A + C ] ) = ( + 1) T E [ K 1 + K + K 3 ] =0 ( + 1) E = E n, n =0 ( + 1) N E = (E 1), and 4 ( + 1) E M = (E 1) 4, =0 it follows that Ψ 5 has the above value Theorem 6 When 0, then the Table 1 gives the values of Ψ 5 In all other cases the product Ψ 5 is equal to λ n (e + f + g + h) Proof of row 1 in Table 1 for Ψ 5 When 0, we have ( + 1) w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) = ( + 1) ( e E = ( + 1) ( 1 [ B1 α a+b + A 1 β a+b ]) ( 1 [ B α c+d c+d + A β ]) + f F + g G + h ) H =0 ( + 1)E = E n, we get Ψ 5 = e E n + ff n + g G n + hh n For any integer n 0, let En = (n + 1) E + 1, En Fn, G n, Hn, Fn, G n and Hn similarly Let = E n (E + 1) n 1 We dene M = n E(E + 1) n 3 (E n + E n E n 1), N = n E(E + 1) n (E n + 1) Theorem 7 (a) When = 0, then T K 3, if n = 0, T E K 4 + K 3 ], if n = 1, Ψ 6 = T [ ] 3 E (K 4 + K + 3 K 1 ) + 4 E K 4 + K 3, if n =, T [ M K 1 + N K + En K 3 ], if n 3 (b) When 0, then Ψ 6 = En Proof (b) Since ( + 1) ( n e + F n f + G n g + Hn h ) w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q) = 1 ( + 1) ( n ) ( e E + f F + g G + h H ),

9 On Sums of Products of Horadam Numbers 491 from n =0 ( + 1) ( ) n E = En, it follows that Ψ 6 indeed has the above value 6 The improved alternating sums, II In this section we study the following sums obtained by multiplying the terms of the sums Ψ 5 and Ψ 6 with the powers of the xed complex number k Of course, for k = 1, we get the sums Ψ 5 and Ψ 6 from the sums Ψ 7 and Ψ 8 Ψ 7 = Ψ 8 = =0 k ( + 1) w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q), =0 ( ) n k ( + 1) w a+b (a 0, b 0 ; p, q) w c+d (c 0, d 0 ; p, q), when a 0, b 0, c 0, d 0, p and q 0 are complex numbers and n 0, a 0, c 0, b > 0 and d > 0 are integers Let E = k α b+d, F = k α b β d, G = k α d β b and H = k β b+d When E 1, for any integer n 0, let E n = (n + 1)En+ (n + )E n We similarly dene (E 1) F n, G n and H n On the other hand, when α b β b, for any integer n 0, we dene b n and b n as in the previous section We similarly dene d n and d n In this section (n + 1)(n + ) λ n is again Theorem 8 The expressions for Ψ 5 in Theorems 5 and 6 describe also the sum Ψ 7 (with the new meaning of E, F, G and H) Theorem 9 The expressions for Ψ 6 in Theorem 7 describe also the sum Ψ 8 (with the new meaning of E, F, G and H) References [1] H Belbachir and F Bencherif, Sums of products of generalized Fibonacci and Lucas numbers, Ars Combinatoria, (to appear) [] Z ƒerin, Sums of products of generalized Fibonacci and Lucas numbers, Demonstratio Mathematica, 4() (009), [3] Z ƒerin, Properties of odd and even terms of the Fibonacci sequence, Demonstratio Mathematica, 39(1) (006), [4] Z ƒerin, On sums of squares of odd and even terms of the Lucas sequence, Proccedings of the Eleventh International Conference on Fibonacci Numbers and their Applications, Congressus Numerantium, 194(009), [5] Z ƒerin, Some alternating sums of Lucas numbers, Central European Journal of Mathematics 3(1) (005), 1-13

10 49 Zvonko čerin [6] Z ƒerin, Alternating Sums of Fibonacci Products, Atti del Seminario Matematico e Fisico dell'università di Modena e Reggio Emilia, 53(005), [7] Z ƒerin and G M Gianella, On sums of squares of Pell-Lucas numbers, INTEGERS: Electronic Journal of Combinatorial Number Theory, 6(006), A15 [8] Z ƒerin and G M Gianella, Formulas for sums of squares and products of Pell numbers, Acc Sc Torino - Atti Sci Fis, 140(006), [9] Z ƒerin and G M Gianella, On sums of Pell numbers, Acc Sc Torino - Atti Sci Fis, 141(007), 3-31 [10] A F Horadam, Generating Functions for Powers of a Certain Generalized Sequence of Numbers, Duke Math J, 3(1965), [11] E Lucas, Théorie des Fonctions Numériques Simplement Périodiques, American Journal of Mathematics 1(1878), [1] N J A Sloane, On-Line Encyclopedia of Integer Sequences,