Journal of the Indian Math. Soc. Vol. 82, Nos. (1 - 2), (2015), 23–32.
ON SUM AND RATIO FORMULAS FOR BALANCING NUMBERS R. K. DAVALA AND G. K. PANDA Abstract. This paper deals with the construction of explicit formulas for sum of consecutive balancing numbers, consecutive even/odd balancing numbers, squares of consecutive balancing numbers, squares of consecutive even/odd balancing numbers and pronic product of balancing numbers. Sums of these numbers with alternative signs give beautiful results. When indices are in arithmetic progression, ratios of sums and differences follow certain interesting patterns.
(Received: February 24, 2014, Accepted: April 8, 2014) 1. Introduction As defined by Behera and second author of this paper in [1], balancing numbers and balancers are solutions of the diophantine equation 1 + 2 + · · · + (n − 1) = (n + 1) + (n + 2) + · · · + (n + r). It is customary to denote the nth balancing number by Bn and the corresponding balancer by Rn . Further, p Cn = 8Bn2 + 1 is called the nth Lucas-balancing number [8], [10, p.25]. The Balancing and Lucas-balancing numbers satisfy the recurrence relations, B1 = 1, B2 = 6, Bn+1 = 6Bn −Bn−1 and C1 = 3, C2 = 17, Cn+1 = 6Cn −Cn−1 , n ≥ 2. The Binet forms (also popularly known as the closed forms) for these numbers are respectively α2n + β 2n α2n − β 2n √ , Cn = Bn = 2 4 2 √ √ where α = 1 + 2 and β = 1 − 2 . Using the Binet forms or rewriting the recurrence relations Bn+1 = 6Bn − Bn−1 and Cn+1 = 6Cn − Cn−1 as Bn−1 = 6Bn − Bn+1 and Cn−1 = 6Cn − Cn+1 , one can easily see that B−n = −Bn and C−n = Cn . 2010 Mathematics Subject Classification. 11B39. Key words and phrases: Balancing and Lucas balancing numbers, Balancer, integer sequences, Binet form. ISSN 0019–5839
c Indian Mathematical Society, 2015 .
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R. K. DAVALA AND G. K. PANDA
The objective of this paper is to develop certain interesting sum formulas involving terms of the sequence of balancing numbers (henceforth we call balancing sequence) in terms of balancing and Lucas balancing numbers. Certain similar sum formulas involving Fibonacci, Lucas and Pell-Lucas numbers are available in [2, 3, 4, 5, 6, 9]. The results contained in the following theorem are very important to the theory of balancing numbers. We will be frequently using these results with or without further reference. Theorem 1.1. For all non-negative integer m, n and x (A) (B) (C) (D) (E)
Bx+2 − Bx = 2Cx+1 , Bm+n = Bm Bn+1 − Bm−1 Bn , Bm±n = Bm Cn ± Cm Bn , Bm±n =3Bm ± Cm , p Rn = 21 −2Bn − 1 + 8Bn2 + 1 = 21 (−2Bn − 1 + Cn ),
(F) (G) (H) (I)
Cx+2 − Cx = 16Bx+1 , Bm = Bx+1 Bm+x − Bx Bm+x+1 , Bm+(n+1)x = 2Bm+nx Cx − Bm+(n−1)x , Pn Bn+2 = 5Bn+1 + 4 i=1 Bi + 1.
Proof. For proofs of (A) to (E), the readers are advised to refer to [7, 10]. One can easily prove (F) and (I) using the recurrence relations of balancing and Lucas-balancing numbers. However, (G) and (H) can be proved as follows: Bx+1 Bm+x − Bx Bm+x+1 = (3Bx + Cx )Bm+x − Bx (3Bm+x + Cm+x ) = Bm+x Cx − Cm+x Bx = Bm , and 2Bm+nx Cx − Bm+(n−1)x = 2Bm+nx Cx − (Bx+1 Bm+nx − Bx Bm+nx+1 ) = Bm+nx (2Cx − Bx+1 ) + Bx Bm+nx+1 = −Bm+nx Bx−1 + Bx Bm+nx+1 = Bm+(n+1)x . 2. Sum formulas with linear and nonlinear terms In this section, we consider certain sum formulas involving linear and nonlinear combinations of balancing numbers. The following theorem deals with the sum of consecutive terms of the balancing sequence with same/alternating signs.
ON SUM AND RATIO FORMULAS FOR BALANCING NUMBERS
25
Theorem 2.1. If k and m are positive integers then m P Bi = 14 (Cm+1 − 2Bm+1 − 1), (a) (b)
(c) (d)
i=1 m P
i=1 m P
i=0 m P
i=0
(−1)i Bi = Bk+2i =
(−1)m (Bm 8
1 16 (Ck+2m+1
+ Bm+1 ) − 81 , − Ck−1 ),
(−1)i Bk+2i = 16 ((−1)m Bk+2m+1 + Bk−1 ).
Proof. (a) By virtue of Theorem 4.1 have m X 1 Bi = Rm+1 = 2 i=1
and 6.2 of [7] and Theorem 1.1 (E), we 1 (Cm+1 − 2Bm+1 − 1). 4
(b) The proof is based on mathematical induction on m. It is easy to see that the assertion is true for m = 1 . Assume that the assertion is true for m = 1, 2, · · · , k. For m = k + 1, we distinguish into two cases : Case I: k is even. In this case, k+1 X
(−1)i Bi =
k X i=1
i=1
(−1)i Bi − Bk+1 =
1 1 (Bk + Bk+1 ) − − Bk+1 8 8
1 1 (Bk − 7Bk+1 − 1) = (−Bk+2 − Bk+1 − 1) 8 8 1 1 = − (Bk+1 + Bk+2 ) − . 8 8
=
Case II: k is odd. In this case, k+1 X
(−1)i Bi =
k X i=1
i=1
1 1 (−1)i Bi + Bk+1 = − (Bk + Bk+1 ) − + Bk+1 8 8
1 1 = − (Bk − 7Bk+1 + 1) = − (−Bk+2 − Bk+1 + 1) 8 8 1 1 = (Bk+1 + Bk+2 ) − . 8 8
In both the cases, the statement is true for m = k + 1, hence by mathematical induction the statement is true for all m. (c) We prove this statement using mathematical induction. For m = 1, 1 (16Bk + 16Bk+2 ) 16 1 1 = (Ck+1 − Ck−1 + Ck+3 − Ck+1 ) = (Ck+3 − Ck−1 ). 16 16
Bk + Bk+2 =
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R. K. DAVALA AND G. K. PANDA
Assume that the assertion is true for m = p, then p+1 X
Bk+2i =
p X
Bk+2i + Bk+2p+2 =
i=0
i=1
=
1 (Ck+2p+1 − Ck−1 ) + Bk+2p+2 16
1 1 (Ck+2p+1 − Ck−1 + 16Bk+2p+2 ) = (Ck+2p+3 − Ck−1 ) 16 16
implies that the statement is true for m = p + 1. By mathematical induction the statement is true for all natural numbers m. (d) Using the Binet form for balancing numbers, we get m X
(−1)i Bk+2i
i=0
m
1 X (−1)i [α2k+4i − β 2k+4i ] = √ 4 2 i=0 " # m m X X 1 = √ α2k (−α4 )i − β 2k (−β 4 )i 4 2 i=0 i=0 1 (−α4 )m+1 − 1 (−β 4 )m+1 − 1 2k 2k = √ α −β −α4 − 1 −β 4 − 1 4 2 m 2k+4m+4 2k (−1)m β 2k+4m+4 + β 2k (−1) α +α 1 − = √ α4 + 1 β4 + 1 4 2 (−1)m α2k+4m+4 + α2k (−1)m β 2k+4m+4 + β 2k 1 − = √ 6α2 6β 2 4 2 1 = √ (−1)m α2k+4m+2 + α2k−2 − (−1)m β 2k+4m+2 − β 2k−2 24 2 1 = {(−1)m Bk+2m+1 + Bk−1 }. 6
The following result is a direct consequence of Theorem 2.1 (d). Corollary 2.1. For any positive integer m,
2m−1 P i=0
(−1)i−1 B2i+1 = 16 B4m .
Remark 2.1. Observe that when k is even/odd, (c) gives sum formulas for consecutive even/odd balancing numbers and (d) gives sum formulas for consecutive even/odd balancing numbers with alternating sign. Further, in view of m−1 m−1 P P −1 2 B1+2i = C2m B1+2i = Bm and by virtue of (c), Theorem 2.2 of [8], 16 . i=0
i=0
2 2 2 Equating the right hand sides, we get C2m = 16Bm + 1 = Cm + 8Bm , which is an alternative proof of corollary 2.4.9 [10, p.27].
ON SUM AND RATIO FORMULAS FOR BALANCING NUMBERS
27
The last theorem provides sum formulas for balancing numbers. Finding sum formulas for squares of balancing numbers is also equally important. In this context, we have the following theorem. Theorem 2.2. For all positive integers k and m, m P 1 2 (B2m+2k+1 − B2k−1 − 2(m + 1)), Bk+i = 32 (a) (b)
i=0 m P
2 (−1)i Bk+i =
i=0 m P
2 (−1)i Bk+2i =
i=0 m P
(c) (d)
i=0
2 Bk+2i =
1 m 96 ((−1) C2m+2k+1
1 192 (B4m+2k+2
+ C2k−1 − 3(1 + (−1)m )),
− B2k−2 − 12(m + 1)),
1 m 544 ((−1) C4m+2k+2
+ C2k−2 − 17(1 + (−1)m )).
Proof. We shall prove (b) and (c) only, as proofs of (a) and (d) are similar. Let k and m be natural numbers. Use of Binet forms for balancing and Lucas balancing numbers gives m X
2 (−1)i Bk+i
i=0
= = = =
=
=
=
m
1 X (−1)i [α2k+2i − β 2k+2i ] 32 i=0
m 1 X (−1)i α4k+4i + β 4k+4i − 2 32 i=0 # " m m m X X X 1 (−1)i (−β 4 )i − 2 (−α4 )i + β 4k α4k 32 i=0 i=0 i=0 1 (−1)m α4k+4m+4 + α4k (−1)m β 4k+4m+4 + β 4k + 32 α4 + 1 β4 + 1 1 − (1 + (−1)m ) 32 (−1)m α4k+4m+4 + α4k (−1)m β 4k+4m+4 + β 4k 1 + 32 6α2 6β 2 1 − (1 + (−1)m ) 32 1 (−1)m α4k+4m+2 + α4k−2 + (−1)m β 4k+4m+2 + β 4k−2 192 1 − (1 + (−1)m ) 32 1 ((−1)m C2m+2k+1 + C2k−1 − 3(1 + (−1)m ) , 96
hence (b) follows. Further since
28
R. K. DAVALA AND G. K. PANDA m X
2 Bk+2i =
i=0
= = = = = =
1 32
m X i=0
[α2k+4i − β 2k+4i ]2
m 1 X 4k+8i α + β 4k+8i − 2 32 i=0 # " m m X X 1 8i 8i 4k 4k β − 2(m + 1) α +β α 32 i=0 i=0 8m+8 8m+8 α −1 β −1 1 4k α4k + β − 2(m + 1) 32 α8 − 1 β8 − 1 4k+8m+8 α − α4k 1 β 4k+8m+8 − β 4k √ √ + − 2(m + 1) 32 24 2α4 24 2β 4 1 1 4k+8m+4 √ α − α4k−4 − β 4k+8m+4 + β 4k−4 − (m + 1) 16 768 2 1 (B4m+2k+2 − B2k−2 − 12(m + 1)) , 192
The proof of (c) is complete.
The following theorem gives the sum formulas for consecutive pronic products of the balancing sequence with same/alternating sign. Theorem 2.3. For any positive integers k and m m P 1 Bk+i Bk+i+1 = 32 (a) (B2m+2k+2 − B2k − 6(m + 1)), (b)
(c) (d)
i=0 m P
i=0 m P
i=0 m P
(−1)i Bk+i Bk+i+1 = Bk+2i Bk+2i+1 =
1 m 96 ((−1) C2m+2k+2
1 192 (B4m+2k+3
(−1)i Bk+2i Bk+2i+1 =
i=0
+ C2k − 9(1 + (−1)m )),
− B2k−1 − 36(m + 1)),
1 m 544 ((−1) C4m+2k+3
+ C2k−1 − 51(1 + (−1)m )).
Proof. We shall prove (b) and (c) only, as the proofs of (a) and (d) are similar. Let k and m be natural numbers. Use of Binet forms for balancing and Lucas balancing numbers gives m X
(−1)i Bk+i Bk+i+1
i=0
m
1 X (−1)i [α4k+4i+2 + β 4k+4i+2 − 6] 32 i=0 # " m m X X 1 4 i m 4 i 4k+2 4k+2 (−β ) − 3(1 + (−1) ) (−α ) + β = α 32 i=0 i=0
=
ON SUM AND RATIO FORMULAS FOR BALANCING NUMBERS
29
1 (−1)m α4m+4 + 1 (−1)m β 4m+4 + 1 4k+2 α4k+2 + β 32 α4 + 1 β4 + 1 3 − (1 + (−1)m ) 32 1 (−1)m α4m+4 + 1 (−1)m β 4m+4 + 1 4k+2 = α4k+2 + β 32 6α2 6β 2 3 − (1 + (−1)m ) 32 1 = (−1)m α4k+4m+4 + α4k + (−1)m β 4k+4m+4 + β 4k 192 3 − (1 + (−1)m ) 32 1 ((−1)m C2m+2k+2 + C2k − 9(1 + (−1)m ) , = 96
=
Thus, (b) follows and (c) follows from m X
Bk+2i Bk+2i+1
i=0
= = = = = =
m
1 X 4k+8i+2 [α + β 4k+8i+2 − 6] 32 i=0 # " m m X X 1 8 i 8 i 4k+2 4k+2 (β ) − 6(m + 1) (α ) + β α 32 i=0 i=0 8m+8 8m+8 1 α −1 β −1 4k+2 α4k+2 + β − 6(m + 1) 32 α8 − 1 β8 − 1 8m+8 8m+8 β −1 1 α −1 √ √ + β 4k+2 − 6(m + 1) α4k+2 32 24 2α4 24 2β 4 6(m + 1) 1 4k+8m+6 √ α − α4k−2 − β 4k+8m+6 + β 4k−2 − 32 768 2 1 (B4m+2k+3 − B2k−1 − 36(m + 1)) . 192
The following theorem provides sum formula involving weighted sum of consecutive balancing numbers. Theorem 2.4. For any positive integer n, nB1 + (n − 1)B2 + · · · + 2Bn−1 + Bn =
1 [Bn+1 − (n + 1)]. 4
Proof. Let us define an integer sequence {Zn } as Zn = nB1 + (n − 1)B2 + · · · + 2Bn−1 + Bn .
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R. K. DAVALA AND G. K. PANDA
We will prove that Zn = 14 [Bn+1 −(n+1)]. The proof is based on mathematical induction. It is easy to see that the assertion is true for n = 1. Assume that the assertion is true for n = k. To complete the proof, we need to show that the assertion is true for n = k + 1. Using Theorem 1.1 (I) and the fact that for n+1 P Bi , we get each natural number n, Zn+1 − Zn = i=1
Zk+1 = Zk +
k+1 X i=1
Bi =
1 1 [Bk+1 − (k + 1)] + [Bk+3 − 5Bk+2 − 1] 4 4
1 1 = [Bk+3 − 5Bk+2 + Bk+1 − (k + 2)] = [Bk+2 − (k + 2)]. 4 4 Thus, the assertion is true for n = k + 1.
3. Formulas involving ratio of linear combinations In this section, we present some quotients involving sums or differences of balancing numbers that simplify to linear expressions of balancing numbers. In some cases, the subscripts of balancing numbers involved in the ratio are in arithmetic progressions. Theorem 3.1. If m and x are natural numbers then each of
Bm+2x+1 ± Bm Bm+x+1 ± Bm+x
Bm+3x ± Bm are independent of m . Also the following identities hold. Bm+2x ± Bm+x Bm+2x+1 − Bm (i) = Bx+1 + Bx , Bm+x+1 − Bm+x Bm+2x+1 + Bm (ii) = Bx+1 − Bx , Bm+x+1 + Bm+x Bm+3x − Bm = Bx+1 − Bx−1 + 1, (iii) Bm+2x − Bm+x Bm+3x + Bm (iv) = Bx+1 − Bx−1 − 1, Bm+2x + Bm+x
and
Proof. By virtue of Theorem 1.1 (B) and (G), we get (Bm+x+1 − Bm+x )(Bx+1 + Bx ) = (Bm+x+1 Bx+1 − Bx+1 Bx ) + (Bx Bm+x+1 − Bx+1 Bm+x ) = B(m+x+1)+x − Bm = Bm+2x+1 − Bm from which (i) follows. Further, (ii) follows from (Bm+x+1 + Bm+x )(Bx+1 − Bx ) = (Bm+x+1 Bx+1 − Bm+x Bx ) + (Bx+1 Bm+x − Bx Bm+x+1 ) = B(m+x+1)+x + Bm = Bm+2x+1 + Bm .
ON SUM AND RATIO FORMULAS FOR BALANCING NUMBERS
31
Using Theorem 1.1 (A), (C) and (H), we get (Bm+2x − Bm+x )(2Cx + 1) = 2Bm+2x Cx − 2Bm+x Cx + Bm+2x − Bm+x = 2Bm+2x Cx − Bm+x Cx + Bx Cm+x − Bm+x = 2Bm+2x Cx − Bm − Bm+x = Bm+3x − Bm and thus (iii) follows and finally, (iv) follows from (Bm+2x + Bm+x )(2Cx − 1) = 2Bm+2x Cx + 2Bm+x Cx − Bm+2x − Bm+x = 2Bm+2x Cx + Bm+x Cx − Bx Cm+x − Bm+x = 2Bm+2x Cx + Bm − Bm+x = Bm+3x + Bm . Theorem 3.2. For natural numbers m, n and k,
Bm+2n+2k − Bm Bn+k = . Bm+n+2k − Bm+n Bk
Proof. It is known that [see 11] if a, b, c, d and r are integers and a + b = c + d, then Ba Bb − Bc Bd = Ba−r Bb−r − Bc−r Bd−r . Thus, Bm+n Bk+n − Bm Bk = Bn Bm+k+n and Bn+k Bm+n+2k − Bk Bm+2n+2k = Bn Bm+k+n . Combining the above identities, we get Bm+n Bk+n − Bm Bk = Bn+k Bm+n+2k − Bk Bm+2n+2k . Rearrangement gives Bm+2n+2k − Bm Bn+k = Bm+n+2k − Bm+n Bk Acknowledgement: The authors are thankful to the anonymous referee for his valuable suggestions which improved the presentation of the paper to a great extent.
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R. K. DAVALA AND G. K. PANDA, Department of Mathematics, National Institute of Technology, Rourkela-769 008, INDIA.
[email protected] gkpanda
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