Finite Reciprocal Sums Involving Summands That are Balanced Products of Generalized

Fibonacci Numbers

R. S. Melham School of Mathematical Sciences University of Technology, Sydney NSW 2007 Australia

                     ray.melham@uts.edu.au

Abstract In this paper we find closed forms, in terms of rational numbers, for certain finite sums. The denominator of each summand is a finite product of terms drawn from two sequences that are generalizations of the Fibonacci and Lucas numbers.

1 Introduction

In [1, 2] we considered certain types of finite reciprocal sums involving generalized Fibonacci numbers. Indeed we gave closed forms, in terms of rational numbers, for these sums. Our purpose here is to give closed forms for finite reciprocal sums that are of a different type than those considered in [1, 2], thereby extending the work in [1, 2]. As in [1, 2], our results can be used to produce finite reciprocal sums that involve the Fibonacci and Lucas numbers.

We begin by introducing the three pairs of integer sequences that feature in this paper. Let a ≥ 0 and b ≥ 0 be integers with (a, b) ≠ (0, 0). For p a positive integer we define, for all integers n, the sequences \{W_{n}\} and \{\overline{W}_{n}\} by

W_{n} = pW_{n−1} + W_{n−2}, W_{0} = a, W_{1} = b,

and

\overline{W}_{n} = W_{n−1} + W_{n+1}.

For (a, b, p) = (0, 1, 1) we have \{W_{n}\}=\{F_{n}\}, and \{\overline{W}_{n}\}=\{L_{n}\}, which are the Fibonacci and Lucas numbers, respectively. Retaining the parameter p, and taking (a, b) = (0, 1), we write \{W_{n}\}=\{U_{n}\}, and \{\overline{W}_{n}\}=\{V_{n}\}, which are integer sequences that generalize the Fibonacci and Lucas numbers, respectively.

Let α and β denote the two distinct real roots of x^{2} − px − 1 = 0. Set A = b − aβ and B = b − aα. Then the closed forms (the Binet forms) for \{W_{n}\} and \{\overline{W}_{n}\} are, respectively,

W_{n} = \frac{Aα^{n} − Bβ^{n}}{α − β} ,

and

\overline{W}_{n} = Aα^{n} + Bβ^{n}.

We require also the constants e_{W} = AB = b^{2} − pab − a^{2}, and Δ = p^{2} + 4.

Throughout this paper we take k ≥ 1, m ≥ 0, and n ≥ 2 to be integers. Let m_{1} < m_{2} and m_{3} < m_{4} be non-negative integers with m_{1} + m_{2} = m_{3} + m_{4}. We begin by giving a closed form for the finite sum

S_{4}(k, m, n, m_{1}, …, m_{4}) = ∑_{i=1}^{n−1} \frac{1}{W_{k(i+m_{1})+m}W_{k(i+m_{2})+m}\overline{W}_{k(i+m_{3})+m}\overline{W}_{k(i+m_{4})+m}} .

Because of the conditions on the m_{i} we consider S_{4} to be the most intriguing sum that we present in this paper. We also give closed forms for similar sums that have longer products in the denominator, and to this end we introduce some notation. For integers 0 < m_{1} < m_{2} write

P_{6}(W, \overline{W} , k, m, i, 0, m_{1}, m_{2})
= W_{ki+m}W_{k(i+m_{1})+m}W_{k(i+m_{2})+m}\overline{W}_{ki+m}\overline{W}_{k(i+m_{1})+m}\overline{W}_{k(i+m_{2})+m}.

Likewise, for integers 0 < m_{1} < m_{2} < m_{3} write

P_{8}(W, \overline{W} , k, m, i, 0, m_{1}, m_{2}, m_{3})
= W_{ki+m}W_{k(i+m_{1})+m} ⋯ W_{k(i+m_{3})+m}\overline{W}_{ki+m}\overline{W}_{k(i+m_{1})+m} ⋯ \overline{W}_{k(i+m_{3})+m}.

We say that P_{8} consists of a balanced product of eight terms that are drawn from the sequences \{W_{n}\} and \{\overline{W}_{n}\}. We also consider such products where the terms are drawn from the sequences {U_{n}} and {V_{n}}, and where the product U_{ki+m}V_{ki+m} is not included. For instance P_{4} (U, V, k, m, i, m_{1}, m_{2}) denotes the product of U_{k(i+m_{1})+m}U_{k(i+m_{2})+m} and V_{k(i+m_{1})+m}V_{k(i+m_{2})+m}. Since U_{n}V_{n} = U_{2n}, P_{4} can be shortened to a product of two terms from the sequence {U_{n}}. However, we choose to retain the longer form for P_{4} (and for expressions analogous P_{4}) in order to highlight the relationship between the numerator and the denominator of the summand. Later, however, when giving examples of our results that involve F_{n} and L_{n}, we present these examples in simplified form.

Using the notation that we have just introduced, we now define three finite sums whose closed forms we give in this paper. Let 0 < m_{1} < m_{2} be integers. Define

S_{6}^{0}(k, m, n, m_{1}, m_{2}) = ∑_{i=1}^{n−1} \frac{P_{2}(U, V, k, m, i, 0)}{P_{6}(W, \overline{W} , k, m, i, 0, m_{1}, m_{2})},

S_{6}^{1}(k, m, n, m_{1}, m_{2}) = ∑_{i=1}^{n−1} \frac{P_{2}(U, V, k, m, i, m_{1})}{P_{6}(W, \overline{W} , k, m, i, 0, m_{1}, m_{2})},

and

S_{6}^{2}(k, m, n, m_{1}, m_{2}) = ∑_{i=1}^{n−1} \frac{P_{2}(U, V, k, m, i, m_{2})}{P_{6}(W, \overline{W} , k, m, i, 0, m_{1}, m_{2})}.

In each of these three cases the numerator of the summand consists of a product of two terms, and the denominator of the summand consists of a product of six terms.

We evaluate each of the finite sums that we consider in this paper in terms of rational numbers. In addition to the four finite sums defined above, we consider analogous finite sums where the denominator of the summand consists of a product of eight, or ten, or twelve terms. In each case the numerator of the summand is either unity, or is a product of terms drawn from the sequences {U_{n}} and {V_{n}} and is defined in terms of the P notation that we have introduced. Furthermore, we consider only finite sums where the number of terms that constitute the product in the denominator of the summand exceeds the number of terms that constitute the product in the numerator of the summand by a multiple of four. Indeed these are the only types of sums, with the structure described earlier in this paragraph, for which we have been able to find closed forms.

In Section 2 we present one result, namely the closed form for S_{4}, and give a proof. In Section 3 we present the closed forms for S_{6}^{0}, S_{6}^{1}, and S_{6}^{2}. In subsequent sections we present a selection of the results that we have found that involve longer products in the denominator of the summand. Indeed, we limit the scope of this paper to finite sums that have four, or six, or eight, or ten, or twelve products in the denominator of the summand.

There are two finite sums that feature throughout. For integers 0 ≤ l_{1} < l_{2} these finite sums are

Ω_{W} (k, m, n, l_{1}, l_{2}) = ∑_{i=l_{1}}^{l_{2}−1} \frac{(−1)^{ki}}{W_{k(i+2)+m}W_{k(i+n)+m}} ,

and

Ω\overline{W} (k, m, n, l_{1}, l_{2}) = ∑_{i=l_{1}}^{l_{2}−1} \frac{(−1)^{ki}}{\overline{W}_{k(i+2)+m}\overline{W}_{k(i+n)+m}} .

To prevent the presentation from becoming too unwieldy, we suppress certain arguments from quantities when there is no danger of confusion. For instance S_{4}(n) will denote S_{4}(k, m, n, m_{1}, m_{2}, m_{3}, m_{4}) when we want n to vary and the other parameters to remain fixed. Likewise Ω_{W} (k, m, n, l_{1}, l_{2}) will be denoted by Ω_{W} (l_{1}, l_{2}) when l_{1} and l_{2} vary and the other parameters remain fixed.

We now give two identities involving Ω_{W} and Ω\overline{W} that are required for the proofs of all the theorems in this paper. We state these identities as lemmas.

Lemma 1. With Ω_{W} as defined above, we have

U_{k(n−1)}Ω_{W} (n + 1) − U_{k(n−2)}Ω_{W} (n) = \frac{(−1)^{k(n+l_{1})}U_{k(l_{2}−l_{1})}}{W_{k(n+l_{1})+m}W_{k(n+l_{2})+m}} .

Lemma 2. With Ω\overline{W} as defined above, we have

U_{k(n−1)}Ω\overline{W} (n + 1) − U_{k(n−2)}Ω\overline{W} (n) = \frac{(−1)^{k(n+l_{1})}U_{k(l_{2}−l_{1})}}{\overline{W}_{k(n+l_{1})+m}\overline{W}_{k(n+l_{2})+m}} .

The proof of Lemma 1 was given in [2], and since the proof of Lemma 2 is similar we refrain from giving it here.

2 A closed form for S_{4}

Let k, m, n, m_{1}, m_{2}, m_{3}, and m_{4} satisfy the constraints given earlier in the definition of S_{4}. Set

a_{0} = a_{0}(k, m_{1}, m_{2}, m_{3}, m_{4}) = e_{W} U_{(m_{4}−m_{3})k}U_{(m_{2}−m_{1})k}V_{(m_{4}−m_{1})k}V_{(m_{4}−m_{2})k}.

Then

Theorem 3. With S_{4} as defined in Section 1,

a_{0} (S_{4}(n) − S_{4}(2)) = (−1)^{m}U_{k(n−2)} ((−1)^{k(m_{1}+m_{3})}U_{(m_{4}−m_{3})k}Ω_{W} (m_{1}, m_{2})
− ΔU_{(m_{2}−m_{1})k}Ω\overline{W} (m_{3}, m_{4})) .

Proof. In [2] we demonstrated two methods of proof, and both methods apply here. The first method required quite a lot to set up, relying heavily upon generalized Fibonacci identities. The second method was more direct, and mechanical, relying upon the closed forms of the relevant sequences. We use the second method here since it is transparent and can be used to effectively prove all the theorems in this paper. To this end, with α = (p + \sqrt{Δ}) /2, it is advantageous to write the closed forms of the sequences in question as

U_{n} = (α^{n} + (−1)^{n+1}α^{−n}) /\sqrt{Δ},
V_{n} = α^{n} + (−1)^{n}α^{−n},
W_{n} = ((b + aα^{−1}) α^{n} + (−1)^{n+1} (b − aα) α^{−n}) /\sqrt{Δ},
\overline{W}_{n} = (b + aα^{−1}) α^{n} + (−1)^{n} (b − aα) α^{−n},

where these closed forms are valid for all integers n. Furthermore we set p = α − α^{−1}, and e_{W} = b^{2} − pab − a^{2}.

We remind the reader that all the finite sums in this paper are defined for n ≥ 2, and so it is for these values of n that the following argument holds. In the statement of Theorem 3, denote the quantity on the left side by L (n) and the quantity on the right side by R (n). We first prove that

L(n + 1) − L(n) = R(n + 1) − R(n).  (1)

With the previously stated restrictions on the relevant parameters, we have

L(n + 1) − L(n) = \frac{e_{W} U_{(m_{4}−m_{3})k}U_{(m_{2}−m_{1})k}V_{(m_{4}−m_{1})k}V_{(m_{4}−m_{2})k}}{W_{k(n+m_{1})+m}W_{k(n+m_{2})+m}\overline{W}_{k(n+m_{3})+m}\overline{W}_{k(n+m_{4})+m}} .  (2)

With the use of Lemma 1 and Lemma 2 we can write down, after some straightforward algebra, the expression for the numerator of R(n + 1) − R(n). This expression is

(−1)^{k(n+m_{3})+m}U_{(m_{4}−m_{3})k}U_{(m_{2}−m_{1})k} (\overline{W}_{k(n+m_{3})+m}\overline{W}_{k(n+m_{4})+m}
−ΔW_{k(n+m_{1})+m}W_{k(n+m_{2})+m}) .  (3)

Furthermore, L(n + 1) − L(n) and R(n + 1) − R(n) have identical denominators, so to prove (1) it is enough to prove that

e_{W} V_{(m_{4}−m_{1})k}V_{(m_{4}−m_{2})k} = (−1)^{k(n+m_{3})+m} (\overline{W}_{k(n+m_{3})+m}\overline{W}_{k(n+m_{4})+m}
−ΔW_{k(n+m_{1})+m}W_{k(n+m_{2})+m}) .  (4)

To this end we consider the difference of the expressions on the left and right sides of (4), replace m_{4} by m_{1} + m_{2} − m_{3}, and express everything in terms of the closed forms. With the use of the computer algebra system Mathematica 8 we find that a factor of the resulting expression is (−1)^{2(k(n+m_{3})+m)} − 1, and this proves (4). This, together with the fact that L(2) = R(2) = 0, establishes Theorem 3.

In the proof above the key identity is (4). Likewise, the proof of each theorem in this paper hinges around the proof of a key identity that is analogous to (4), and each such identity follows immediately by substitution of the appropriate closed forms. The method is mechanical and is not dependent upon and special identities. However, the use of a computer algebra system (in our case Mathematica 8) is essential. The proof above serves as a template for the proof of each theorem in this paper, and so we state the theorems in the sections that follow without proof.

We pause to give two examples. Let k = 1, m = 0, and (m_{1}, m_{2}, m_{3}, m_{4}) = (0, 3, 1, 2). Then for W_{n} = F_{n} the result in Theorem 3 becomes

36 ∑_{i=1}^{n−1} \frac{1}{F_{i}F_{i+3}L_{i+1}L_{i+2}} = 1 + F_{n−2} (\frac{6}{F_{n}} − \frac{3}{F_{n+1}} + \frac{2}{F_{n+2}} − \frac{15}{L_{n+1}} ) .

Next let k = 2, m = 0, and (m_{1}, m_{2}, m_{3}, m_{4}) = (2, 3, 1, 4). Then for W_{n} = F_{n} the result in Theorem 3 becomes

40790736 ∑_{i=1}^{n−1} \frac{1}{F_{2(i+2)}F_{2(i+3)}L_{2(i+1)}L_{2(i+4)}}

= 282 + F_{2(n−2)} (\frac{92496}{F_{2(n+2)}} − \frac{67445}{L_{2(n+1)}} − \frac{25830}{L_{2(n+2)}} − \frac{9870}{L_{2(n+3)}} ) .

3 Closed forms for S_{6}^{0}, S_{6}^{1}, and S_{6}^{2}

As stated in the introduction, here, and in the sequel we take k ≥ 1, m ≥ 0, and n ≥ 2 to be integers. In this section we take 0 < m_{1} < m_{2} to be integers.

For 0 ≤ i ≤ 2 define a_{i} = a_{i}(k, m, m_{1}, m_{2}) as

a_{0} = 2(−1)^{m+1}e_{W}^{3} U_{2m_{1}k}U_{2m_{2}k}U_{2(m_{2}−m_{1})k},
a_{1} = U_{2(m_{2}−m_{1})k}W_{0}\overline{W}_{0},
a_{2} = −U_{2m_{1}k}W_{m_{2}k}\overline{W}_{m_{2}k}.

We then have

Theorem 4. With S_{6}^{0} as defined in Section 1,

a_{0} (S_{6}^{0}(n) − S_{6}^{0}(2)) = U_{k(n−2)} (a_{1}Ω_{W} (0, m_{1}) + a_{2}Ω_{W} (m_{1}, m_{2})
− Δa_{1}Ω\overline{W} (0, m_{1}) − Δa_{2}Ω\overline{W} (m_{1}, m_{2})) .

Next, for 1 ≤ i ≤ 2 define b_{i} = b_{i}(k, m_{1}, m_{2}) by

b_{1} = U_{2(m_{2}−m_{1})k}W_{−m_{1}k}\overline{W}_{−m_{1}k},
b_{2} = −U_{2m_{1}k}W_{(m_{2}−m_{1})k}\overline{W}_{(m_{2}−m_{1})k}.

With a_{0} as for Theorem 4, we have

Theorem 5. Let S_{6}^{1} be as defined in Section 1. Then

a_{0} (S_{6}^{1}(n) − S_{6}^{1}(2)) = U_{k(n−2)} (b_{1}Ω_{W} (0, m_{1}) + b_{2}Ω_{W} (m_{1}, m_{2})
− Δb_{1}Ω\overline{W} (0, m_{1}) − Δb_{2}Ω\overline{W} (m_{1}, m_{2})) .

For 1 ≤ i ≤ 2 define c_{i} = c_{i}(k, m_{1}, m_{2}) as

c_{1} = U_{2(m_{2}−m_{1})k}W_{−m_{2}k}\overline{W}_{−m_{2}k},
c_{2} = −U_{2m_{1}k}W_{0}\overline{W}_{0}.

Then, with a_{0} as for Theorem 4, we have Theorem 6. Let S_{6}^{2} be as defined in Section 1. Then

a_{0} (S_{6}^{2}(n) − S_{6}^{2}(2)) = U_{k(n−2)} (c_{1}Ω_{W} (0, m_{1}) + c_{2}Ω_{W} (m_{1}, m_{2})
− Δc_{1}Ω\overline{W} (0, m_{1}) − Δc_{2}Ω\overline{W} (m_{1}, m_{2})) .

Let k = 1, m = 0, and (m_{1}, m_{2}) = (1, 2). Take W_{n} = F_{n+2}. Then with the use of the identity F_{n}L_{n} = F_{2n} to simplify the summand, and also the right side, the result in Theorem 5 becomes

27720 ∑_{i=1}^{n−1} \frac{F_{2(i+1)}}{F_{2(i+2)}F_{2(i+3)}F_{2(i+4)}} = 9 + 8F_{2(n−2)} (\frac{−55}{F_{2(n+2)}} + \frac{168}{F_{2(n+3)}} ) .

4 The summand has eight factors in the denominator

In this section we take 0 < m_{1} < m_{2} < m_{3} to be integers.

Define the finite sum

S_{8}(k, m, n, m_{1}, m_{2}, m_{3}) = ∑_{i=1}^{n−1} \frac{1}{P_{8}(W, \overline{W} , k, m, i, 0, m_{1}, m_{2}, m_{3})}.

For 0 ≤ i ≤ 3 define a_{i} = a_{i}(k, m, m_{1}, m_{2}, m_{3}) by

a_{0} = 2e_{W}^{3} U_{2m_{1}k}U_{2m_{2}k}U_{2m_{3}k}U_{2(m_{3}−m_{2})k}U_{2(m_{3}−m_{1})k}U_{2(m_{2}−m_{1})k},
a_{1} = (−1)^{m}U_{2(m_{3}−m_{2})k}U_{2(m_{3}−m_{1})k}U_{2(m_{2}−m_{1})k},
a_{2} = (−1)^{m+1}U_{2m_{1}k}U_{2(m_{3}−m_{2})k}U_{2(m_{3}+m_{2}−m_{1})k},
a_{3} = (−1)^{m}U_{2m_{1}k}U_{2m_{2}k}U_{2(m_{2}−m_{1})k}.

Then

Theorem 7. We have

a_{0} (S_{8}(n) − S_{8}(2)) = U_{k(n−2)} (a_{1}Ω_{W} (0, m_{1}) + a_{2}Ω_{W} (m_{1}, m_{2})
+ a_{3}Ω_{W} (m_{2}, m_{3}) − Δa_{1}Ω\overline{W} (0, m_{1})
− Δa_{2}Ω\overline{W} (m_{1}, m_{2}) − Δa_{3}Ω\overline{W} (m_{2}, m_{3})) .

We have found that as the number of products in the denominator of the summand increases it becomes more difficult to write down the closed form of the corresponding finite sum. The same is true as the number of products in the numerator of the summand increases. Indeed, to find the closed form for the finite sum T_{8}, defined below, we needed to specialize the values of m_{1}, m_{2}, and m_{3} in a manner that we soon make clear.

Define the sum

T_{8}(k, m, n, m_{1}, m_{2}, m_{3}) = ∑_{i=1}^{n−1} \frac{P_{4} (U, V, k, m, i, m_{1}, m_{2})}{P_{8}(W, \overline{W} , k, m, i, 0, m_{1}, m_{2}, m_{3})}.

Here, and in the sequel, we take g ≥ 1 to be an integer. For 0 ≤ i ≤ 3 define the quantities b_{i} = b_{i}(g, k, m) as follows:

b_{0} = 2e_{W}^{5} U_{2gk}U_{4gk}U_{6gk},
b_{1} = (−1)^{m}W_{−2gk}W_{−gk}\overline{W}_{−2gk}\overline{W}_{−gk},
b_{2} = (−1)^{m+1}V_{2gk}W_{−gk}W_{gk}\overline{W}_{−gk}\overline{W}_{gk},
b_{3} = (−1)^{m}W_{gk}W_{2gk}\overline{W}_{gk}\overline{W}_{2gk}.

We now state our next theorem.

Theorem 8. Let (m_{1}, m_{2}, m_{3}) = (g, 2g, 3g), so that 0, m_{1}, m_{2}, and m_{3} form an arithmetic progression. Then

b_{0} (T_{8}(n) − T_{8}(2)) = U_{k(n−2)} (b_{1}Ω_{W} (0, g) + b_{2}Ω_{W} (g, 2g)
+ b_{3}Ω_{W} (2g, 3g) − Δb_{1}Ω\overline{W} (0, g)
− Δb_{2}Ω\overline{W} (g, 2g) − Δb_{3}Ω\overline{W} (2g, 3g)) .

Let k = 1, m = 0, and g = 1. Then for W_{n} = F_{n+3} the result in Theorem 8 becomes

167207040 ∑_{i=1}^{n−1} \frac{F_{2(i+1)}F_{2(i+2)}}{F_{2(i+3)}F_{2(i+4)}F_{2(i+5)}F_{2(i+6)}}

= 64 + 63F_{2(n−2)} (\frac{6032}{F_{2(n+3)}} − \frac{145145}{F_{2(n+4)}} + \frac{338800}{F_{2(n+5)}}) .

We attempted to find the closed forms for certain finite sums analogous to T_{8}, but without success. Firstly, we considered T_{8} as defined above but with the m_{i} defined as different multiples of g, such as (m_{1}, m_{2}, m_{3}) = (2g, 3g, 5g). Secondly, in the definition of T_{8}, we replaced P_{4} (U, V, k, m, m_{1}, m_{2}) by P_{4} (U, V, k, m, 0, m_{2}), and by P_{4} (U, V, k, m, m_{1}, m_{3}). In attempting to find the corresponding closed forms in each of these two cases we set (m_{1}, m_{2}, m_{3}) = (g, 2g, 3g). Put simply, there seems to be a fine line between success and failure.

5 The summand has ten factors in the denominator

In this section we take 0 < m_{1} < m_{2} < m_{3} < m_{4} to be integers.

Define the sum

S_{10}(k, m, n, m_{1}, m_{2}, m_{3}, m_{4}) = ∑_{i=1}^{n−1} \frac{P_{6} (U, V, k, m, i, m_{1}, m_{2}, m_{3})}{P_{10}(W, \overline{W} , k, m, i, 0, m_{1}, m_{2}, m_{3}, m_{4})}.

For 0 ≤ i ≤ 4 define the quantities a_{i} = a_{i}(g, k, m) as follows:

a_{0} = 2e_{W}^{7} U_{2gk}U_{4gk}U_{6gk}U_{8gk},
a_{1} = (−1)^{m+1}W_{−3gk}W_{−2gk}W_{−gk}\overline{W}_{−3gk}\overline{W}_{−2gk}\overline{W}_{−gk},
a_{2} = (−1)^{m} (V_{4gk} + 1) W_{−2gk}W_{−gk}W_{gk}\overline{W}_{−2gk}\overline{W}_{−gk}\overline{W}_{gk},
a_{3} = (−1)^{m+1} (V_{4gk} + 1) W_{−gk}W_{gk}W_{2gk}\overline{W}_{−gk}\overline{W}_{gk}\overline{W}_{2gk},
a_{4} = (−1)^{m}W_{gk}W_{2gk}W_{3gk}\overline{W}_{gk}\overline{W}_{2gk}\overline{W}_{3gk}.

Once again, to discover our next result we needed the m_{i} to take on special values.

Theorem 9. Let (m_{1}, m_{2}, m_{3}, m_{4}) = (g, 2g, 3g, 4g), so that 0, m_{1}, m_{2}, m_{3}, and m_{4} form an arithmetic progression. Then

a_{0} (S_{10}(n) − S_{10}(2)) = U_{k(n−2)} (a_{1}Ω_{W} (0, g) + a_{2}Ω_{W} (g, 2g)
+ a_{3}Ω_{W} (2g, 3g) + a_{4}Ω_{W} (3g, 4g)
− Δa_{1}Ω\overline{W} (0, g) − Δa_{2}Ω\overline{W} (g, 2g)
− Δa_{3}Ω\overline{W} (2g, 3g) − Δa_{4}Ω\overline{W} (3g, 4g)) .

We have found that the most interesting examples of our results occur when we take W_{n} = F_{n+c} for some non-negative integer c. Accordingly, as an instance of Theorem 9 let k = 1, m = 0, and g = 1. Then for W_{n} = F_{n+5} Theorem 9 yields

b_{0} ∑_{i=1}^{n−1} \frac{F_{2(i+1)}F_{2(i+2)}F_{2(i+3)}}{F_{2(i+5)}F_{2(i+6)}F_{2(i+7)}F_{2(i+8)}F_{2(i+9)}}

= 7 + 144F_{2(n−2)} (\frac{b_{1}}{F_{2(n+5)}} + \frac{b_{2}}{F_{2(n+6)}} + \frac{b_{3}}{F_{2(n+7)}} + \frac{b_{4}}{F_{2(n+8)}} ) ,

where

b_{0} = 13009146630480,
b_{1} = −239632085,
b_{2} = 35147981440,
b_{3} = −632668766730,
b_{4} = 1419740509642.

We also considered variants of S_{10} that we obtained in ways similar to those described in the paragraph at the end of Section 4. However, in each case we were unable to find the closed form of the corresponding finite sum. Furthermore, we considered summands with only two factors in the numerator. Once again, in each case, we were unable to find the closed form of the corresponding finite sum. Theorem 9 is the only result of its kind (i.e., where the summand has ten factors in the denominator, and where the m_{i} are multiples of a positive integer parameter g) that we could find. We have discovered closed forms for such sums where the m_{i} are specific integers, but we refrain from giving these sums.

6 The summand has twelve factors in the denominator

In this section we take 0 < m_{1} < m_{2} < m_{3} < m_{4} < m_{5} to be integers.

Define the sum

S_{12}(k, m, n, m_{1}, m_{2}, m_{3}, m_{4}, m_{5}) = ∑_{i=1}^{n−1} \frac{1}{P_{12}(W, \overline{W} , k, m, i, 0, m_{1}, m_{2}, m_{3}, m_{4}, m_{5})}.

For 0 ≤ i ≤ 5 we define the quantities a_{i} = a_{i}(g, k, m) as

a_{0} = 2e_{W}^{5} U_{2gk}U_{4gk}U_{8gk}U_{10gk}U_{12gk},
a_{1} = a_{5} = (−1)^{m},
a_{2} = a_{4} = (−1)^{m+1} (V_{8gk} + V_{4gk} − 1) ,
a_{3} = (−1)^{m} (V_{12gk} − V_{8gk} + 2) .

In our next theorem 0, m_{1}, m_{2}, m_{3}, m_{4}, and m_{5} are not required to form an arithmetic progression.

Theorem 10. Setting (m_{1}, m_{2}, m_{3}, m_{4}, m_{5}) = (g, 2g, 4g, 5g, 6g) we have

a_{0} (S_{12}(n) − S_{12}(2)) = U_{k(n−2)} (a_{1}Ω_{W} (0, g) + a_{2}Ω_{W} (g, 2g)
+ a_{3}Ω_{W} (2g, 4g) + a_{4}Ω_{W} (4g, 5g)
+ a_{5}Ω_{W} (5g, 6g) − Δa_{1}Ω\overline{W} (0, g)
− Δa_{2}Ω\overline{W} (g, 2g) − Δa_{3}Ω\overline{W} (2g, 4g)
− Δa_{4}Ω\overline{W} (4g, 5g) − Δa_{5}Ω\overline{W} (5g, 6g)) .

We also managed to find a closed form for S_{12} if (m_{1}, m_{2}, m_{3}, m_{4}, m_{5}) = (g, 2g, 3g, 4g, 5g), but we refrain from giving this result here.

To indicate other types of results that are possible, define the sum

T_{12}(k, m, n, m_{1}, m_{2}, m_{3}, m_{4}, m_{5}) = ∑_{i=1}^{n−1} \frac{P_{8} (U, V, k, m, i, m_{1}, m_{2}, m_{3}, m_{4})}{P_{12}(W, \overline{W} , k, m, i, 0, m_{1}, m_{2}, m_{3}, m_{4}, m_{5})}.

We managed to find a closed form for T_{12} under the assumption that the m_{i} are certain multiples of g. One such instance is for (m_{1}, m_{2}, m_{3}, m_{4}, m_{5})=(g, 2g, 3g, 4g, 5g). Another instance is for (m_{1}, m_{2}, m_{3}, m_{4}, m_{5})=(2g, 3g, 4g, 5g, 7g). We have also discovered other results of a similar nature that we do not present here.

7 Concluding comments

Earlier we stated that we chose to limit the scope of this paper to finite sums that have four, or six, or eight, or ten, or twelve products in the denominator of the summand. We have, however, discovered closed forms for finite sums (with the structure described in the introduction) that have fourteen, or sixteen, or eighteen, or twenty products in the denominator of the summand. The possibilities seem limitless.

References

[1] R. S. Melham, Finite sums that involve reciprocals of products of generalized Fibonacci numbers, Integers 13 (2013), #A40. Available at

   http://www.integers-ejcnt.org/vol13.html.

[2] R. S. Melham, More on finite sums that involve reciprocals of products of generalized Fibonacci numbers, Integers 14 (2014), #A4. Available at

   http://www.integers-ejcnt.org/vol14.html.

2010 Mathematics Subject Classification: Primary 11B39; Secondary 11B37. Keywords: reciprocal summation, Fibonacci number, Lucas number.

(Concerned with sequences \underline{A000045} and \underline{A000032}.)

Received February 4 2014; revised version received April 16 2014. Published in Journal of Integer Sequences, May 12 2014.

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