Factored Closed-form Expressions for the Sums of Cubes of Fibonacci and Lucas Numbers

Kunle Adegoke Department of Physics and Engineering Physics Obafemi Awolowo University 220005 Ile-Ife Nigeria

                      adegoke00@gmail.com

Abstract We obtain factored closed-form expressions for the sums of cubes of Fibonacci and

Lucas numbers.

1 Introduction

The Fibonacci numbers, F_{n}, and Lucas numbers, L_{n}, are defined, for n ∈ ℤ, as usual, through the recurrence relations F_{n} = F_{n−1} + F_{n−2}, F_{0} = 0, F_{1} = 1 and L_{n} = L_{n−1} + L_{n−2}, L_{0} = 2, L_{1} = 1, with F_{−n} = (−1)^{n−1}F_{n} and L_{−n} = (−1)^{n}L_{n}.

Clary and Hemenway [2] derived the remarkable formulas

4 ∑_{k=1}^{n} F_{2k}^{3} = \begin{cases}F_{n}^{2}L_{n+1}^{2}F_{n−1}L_{n+2}, & if n is even; \\ L_{n}^{2}F_{n+1}^{2}L_{n−1}F_{n+2}, & if n is odd, \end{cases}  (1)

and

8 ∑_{k=1}^{n} F_{4k}^{3} = F_{2n}^{2}F_{2n+2}^{2}(L_{4n+2} + 6) .  (2)

In this present paper we will derive the following corresponding Lucas counterparts of (1) and (2):

4 ∑_{k=1}^{n} L_{2k}^{3} = \begin{cases}5F_{n}F_{n+1}(L_{n}L_{n+1}L_{2n+1} + 16), & if n is even; \\ L_{n}L_{n+1}(5F_{n}F_{n+1}L_{2n+1} + 16), & if n is odd, \end{cases}  (3)

and

8 ∑_{k=1}^{n} L_{4k}^{3} = F_{2n}L_{2n+2}(5L_{2n}F_{2n+2}F_{4n+2} + 32) .  (4)

In fact we will derive the following more general results:

• If r is odd, then

L_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = \begin{cases}F_{rn}^{2}L_{rn+r}^{2}(L_{rn}F_{rn+r} + F_{r}), & if n is even; \\ L_{rn}^{2}F_{rn+r}^{2}(F_{rn}L_{rn+r} + F_{r}), & if n is odd, \end{cases}  (5)

and

L_{3r} ∑_{k=1}^{n} L_{2rk}^{3} = \begin{cases}5F_{rn}F_{rn+r}(L_{rn}L_{rn+r}L_{2rn+r} + 4(L_{2r} + 1)), & if n is even; \\ L_{rn}L_{rn+r}(5F_{rn}F_{rn+r}L_{2rn+r} + 4(L_{2r} + 1)), & if n is odd. \end{cases} (6)

• If r is even, then

F_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = F_{rn}^{2}F_{rn+r}^{2}(L_{rn}L_{rn+r} + L_{r})  (7)

and

F_{3r} ∑_{k=1}^{n} L_{2rk}^{3} = F_{rn}L_{rn+r}(5L_{rn}F_{rn+r}F_{2rn+r} + 4(L_{2r} + 1)) .  (8)

As variations on identities (5) and (7) we will prove

• If r is odd, then

L_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = \begin{cases}F_{rn}L_{rn+r}(L_{rn}F_{rn+r}F_{2rn+r} − 2F_{r}^{2}), & if n is even; \\ L_{rn}F_{rn+r}(F_{rn}L_{rn+r}F_{2rn+r} − 2F_{r}^{2}), & if n is odd. \end{cases}

• If r is even, then

5F_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = F_{rn}F_{rn+r}(L_{rn}L_{rn+r}L_{2rn+r} − 2L_{r}^{2}) .

2 Required identities and preliminary results

2.1 Telescoping summation identity

The following telescoping summation identity is a special case of more general identities proved by Adegoke [1]. Lemma 1. If f(k) is a real sequence and m, q and n are positive integers, then

∑_{k=1}^{n} [f(mk + mq) − f(mk)] = ∑_{k=1}^{q} f(mk + mn) − ∑_{k=1}^{q} f(mk) .

2.2 First-power Fibonacci summation identities

Lemma 2. If r and n are integers, then

(i) If r is even, then

F_{r} ∑_{k=1}^{n} F_{2rk} = F_{rn}F_{rn+r} .

(ii) If r is odd, then

L_{r} ∑_{k=1}^{n} F_{2rk} = \begin{cases}F_{rn}L_{rn+r}, & if n is even; \\ L_{rn}F_{rn+r}, & if n is odd. \end{cases}

Proof. Setting v = 2r and u = 2rk in the identity

L_{u+v} − (−1)^{v}L_{u−v} = 5F_{u}F_{v}  (9)

gives

L_{2rk+2r} − L_{2rk−2r} = 5F_{2r}F_{2rk} .  (10)

Taking f(k) = L_{k−2r}, q = 2 and m = 2r in Lemma 1 and employing identity (10) we have

\begin{aligned}5F_{2r} ∑_{k=1}^{n} F_{2rk} & = ∑_{k=1}^{2} L_{2rk+2rn−2r} − ∑_{k=1}^{2} L_{2rk−2r} \\ & = L_{2rn+2r} + L_{2rn} − L_{2r} − 2 . \end{aligned}  (11)

If r is even, then on account of the identity

L_{u+v} + (−1)^{v}L_{u−v} = L_{u}L_{v},  (12)

we have

L_{2rn+2r} + L_{2rn} = L_{r}L_{2rn+r}, L_{2r} + 2 = L_{r}^{2} ,

and since

F_{2u} = F_{u}L_{u} ,  (13)

identity (11) now becomes

\begin{aligned}5F_{r} ∑_{k=1}^{n} F_{2rk} & = L_{2rn+r} − L_{r} \\ & = 5F_{rn}F_{rn+r} , by (9) , \end{aligned}  (14)

that is,

F_{r} ∑_{k=1}^{n} F_{2rk} = F_{rn}F_{rn+r}, r even ,

and the first part of Lemma 2 is proved.

If r is odd, then on account of the identities (9) and (12), we have

L_{2rn+2r} + L_{2rn} = 5F_{r}F_{2rn+r}, L_{2r} + 2 = 5F_{r}^{2} ,

and identity (11) reduces to

L_{r} ∑_{k=1}^{n} F_{2rk} = F_{2rn+r} − F_{r}

= \begin{cases}F_{rn}L_{rn+r}, & if n is even; \\ L_{rn}F_{rn+r}, & if n is odd, \end{cases}

and the second part of Lemma 2 is proved. In the last stage of the above derivation we made use of the identities

F_{u+v} − (−1)^{v}F_{u−v} = F_{v}L_{u}  (15)

and

F_{u+v} + (−1)^{v}F_{u−v} = L_{v}F_{u} .  (16)

2.3 First-power Lucas summation identities

Lemma 3. If r and n are integers, then

(i) If r is even, then

F_{r} ∑_{k=1}^{n} L_{2rk} = F_{rn}L_{rn+r} .

(ii) If r is odd, then

L_{r} ∑_{k=1}^{n} L_{2rk} = \begin{cases}5F_{rn}F_{rn+r}, & if n is even; \\ L_{rn}L_{rn+r}, & if n is odd. \end{cases}

Proof. Setting v = 2r and u = 2rk in the identity (15) gives

F_{2rk+2r} − F_{2rk−2r} = F_{2r}L_{2rk} .  (17)

Taking f(k) = F_{k−2r}, q = 2 and m = 2r in Lemma 1 and employing identity (17) we have

\begin{aligned}F_{2r} ∑_{k=1}^{n} L_{2rk} & = ∑_{k=1}^{2} F_{2rk+2rn−2r} − ∑_{k=1}^{2} F_{2rk−2r} \\ & = F_{2rn+2r} + F_{2rn} − F_{2r} . \end{aligned}  (18)

If r is even, then choosing v = r and u = 2rn + r in identity (16) gives

F_{2rn+2r} + F_{2rn} = L_{r}F_{2rn+r}  (19)

and, on account of identity (13), the identity (18) reduces to

F_{r} ∑_{k=1}^{n} L_{2rk} = F_{2rn+r} − F_{r}
= F_{rn+r+rn} − F_{rn+r−rn}
= F_{rn}L_{rn+r} , by identity (15),

and the first part of Lemma 3 is proved.

If r is odd, then choosing v = r and u = 2rn + r in identity (15) gives

F_{2rn+2r} + F_{2rn} = F_{r}L_{2rn+r}  (20)

and, again on account of identity (13), the identity (18) now reduces to

L_{r} ∑_{k=1}^{n} L_{2rk} = L_{2rn+r} − L_{r}

= L_{rn+r+rn} − L_{rn+r−rn}

= \begin{cases}5F_{rn}F_{rn+r}, & if n is even; \\ L_{rn}L_{rn+r}, & if n is odd, \end{cases}

where in the last step we used the identities (9) and (12).

2.4 Other identities

Lemma 4. If r and n are integers, then

\frac{F_{3rn}F_{3rn+3r}}{F_{rn}F_{rn+r}} = L_{rn}L_{rn+r}L_{2rn+r} + L_{2r} + (−1)^{r−1}.

Proof. Using the identity Clary [2, Eq. (36)], or Dresel [3, Eq. (3.3)], namely,

F_{3u} = 5F_{u}^{3} + 3(−1)^{u}F_{u} ,  (21)

we have

\begin{aligned}\frac{F_{3rn}F_{3rn+3r}}{F_{rn}F_{rn+r}} & = (5F_{rn}^{2} + 3(−1)^{rn})(5F_{rn+r}^{2} + 3(−1)^{rn+r}) \\ & = (L_{rn}^{2} − (−1)^{rn})(L_{rn+r}^{2} − (−1)^{rn+r}) \\ & = L_{rn}^{2}L_{rn+r}^{2} − (−1)^{rn+r}L_{rn}^{2} − (−1)^{rn}L_{rn+r}^{2} + (−1)^{r} , \end{aligned}  (22)

where we have also made use of the identity

5F_{u}^{2} − L_{u}^{2} = (−1)^{u−1}4 .  (23)

Now,

L_{rn}^{2}L_{rn+r}^{2} = L_{rn}L_{rn+r}(L_{rn}L_{rn+r})
= L_{rn}L_{rn+r}(L_{2rn+r} + (−1)^{rn}L_{r}) by (12)
= L_{rn}L_{rn+r}L_{2rn+r} + (−1)^{rn}L_{rn}L_{rn+r}L_{r} .

Therefore

\frac{F_{3rn}F_{3rn+3r}}{F_{rn}F_{rn+r}} = L_{rn}L_{rn+r}L_{2rn+r} + (−1)^{rn}L_{rn+r}(L_{rn}L_{r} − L_{rn+r}) − (−1)^{rn+r}L_{rn}^{2} + (−1)^{r} .

But

(−1)^{rn}L_{rn+r}(L_{rn}L_{r} − L_{rn+r})
= (−1)^{rn}L_{rn+r}(L_{rn+r} + (−1)^{r}L_{rn−r} − L_{rn+r}), by (12)
= (−1)^{rn+r}L_{rn+r}L_{rn−r}
= (−1)^{rn+r}(L_{2rn} + (−1)^{rn−r}L_{2r}), again by (12)
= (−1)^{rn+r}L_{2rn} + L_{2r} .

Thus

\frac{F_{3rn}F_{3rn+3r}}{F_{rn}F_{rn+r}} = L_{rn}L_{rn+r}L_{2rn+r} + (−1)^{rn+r}L_{2rn} + L_{2r} − (−1)^{rn+r}L_{rn}^{2} + (−1)^{r}
= L_{rn}L_{rn+r}L_{2rn+r} + (−1)^{rn+r}(L_{2rn} − L_{rn}^{2}) + L_{2r} + (−1)^{r} .

Finally, using the identity

L_{2u} = L_{u}^{2} + (−1)^{u−1}2 ,  (24)

obtained by setting v = u in identity (12), we have the statement of the Lemma.

Lemma 5. If r and n are integers, then

\frac{L_{3rn}L_{3rn+3r}}{L_{rn}L_{rn+r}} = 5F_{rn}F_{rn+r}L_{2rn+r} + L_{2r} + (−1)^{r−1}.

Proof. Using the following identity, of Dresel [3, Eq. (1.6)]

L_{3u} = L_{u}^{3} − 3(−1)^{u}L_{u} ,  (25)

we have

\frac{L_{3rn}L_{3rn+3r}}{L_{rn}L_{rn+r}} = (L_{rn}^{2} − 3(−1)^{rn})(L_{rn+r}^{2} − 3(−1)^{rn+r})
= (5F_{rn}^{2} + (−1)^{rn})(5F_{rn+r}^{2} + (−1)^{rn+r}), by (23)
= 25F_{rn}^{2}F_{rn+r}^{2} + (−1)^{rn+r}5F_{rn}^{2} + (−1)^{rn}5F_{rn+r}^{2} + (−1)^{r} ,

and the rest of the calculation then proceeds as in the proof of Lemma 4, the basic required identities now being (9), (16) and the identity

L_{2u} = 5F_{u}^{2} + (−1)^{u}2,  (26)

obtained by setting v = u in identity (9). Lemma 6. If r and n are integers, then

\frac{L_{3rn}F_{3rn+3r}}{L_{rn}F_{rn+r}} = 5F_{rn}L_{rn+r}F_{2rn+r} + L_{2r} + (−1)^{r}.

Lemma 7. If r and n are integers, then

\frac{F_{3rn}L_{3rn+3r}}{F_{rn}L_{rn+r}} = 5L_{rn}F_{rn+r}F_{2rn+r} + L_{2r} + (−1)^{r}.

Different but equivalent versions of Lemmas 4–7 are given below: Lemma 8. If r and n are integers, then

\frac{F_{3rn}F_{3rn+3r}}{F_{rn}F_{rn+r}} = L_{2rn+r}^{2} + (−1)^{nr}L_{rn+r}^{2} + (−1)^{(n−1)r}L_{rn}^{2} + L_{r}^{2} + (−1)^{r−1}7 .

Proof. The proof is similar to that of Lemma 4, but here we use

L_{rn}^{2}L_{rn+r}^{2} = (L_{2rn+r} + (−1)^{rn}L_{r})^{2}
= L_{2rn+r}^{2} + L_{r}^{2} + 2(−1)^{rn}(L_{r}L_{2rn+r})
= L_{2rn+r}^{2} + L_{r}^{2} + 2(−1)^{rn}(L_{2rn+2r} + (−1)^{r}L_{2rn})

= L_{2rn+r}^{2} + L_{r}^{2} + 2(−1)^{rn}(L_{rn+r}^{2} + (−1)^{rn+r−1}2 + (−1)^{r}(L_{rn}^{2} + (−1)^{rn−1}2)),

and substitute in (22).

Lemma 9. If r and n are integers, then

\frac{L_{3rn}L_{3rn+3r}}{L_{rn}L_{rn+r}} = L_{2rn+r}^{2} + (−1)^{nr−1}L_{rn+r}^{2} − (−1)^{(n−1)r}L_{rn}^{2} + L_{r}^{2} + (−1)^{r} .

Lemma 10. If r and n are integers, then

\frac{L_{3rn}F_{3rn+3r}}{L_{rn}F_{rn+r}} = 5F_{2rn+r}^{2} + (−1)^{nr−1}5F_{rn+r}^{2} + (−1)^{(n−1)r}5F_{rn}^{2} + 5F_{r}^{2} + (−1)^{r}3 .

Lemma 11. If r and n are integers, then

\frac{F_{3rn}L_{3rn+3r}}{F_{rn}L_{rn+r}} = 5F_{2rn+r}^{2} + (−1)^{nr}5F_{rn+r}^{2} − (−1)^{(n−1)r}5F_{rn}^{2} + 5F_{r}^{2} + (−1)^{r}3 .

3 Main results

3.1 Sums of cubes of Fibonacci numbers

Theorem 12. If r and n are integers such that r is odd, then

L_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = \begin{cases}F_{rn}L_{rn+r}(L_{rn}F_{rn+r}F_{2rn+r} − 2F_{r}^{2}), & if n is even; \\ L_{rn}F_{rn+r}(F_{rn}L_{rn+r}F_{2rn+r} − 2F_{r}^{2}), & if n is odd. \end{cases}

Proof. Setting u = 2rk in identity (21) and summing, we have

5 ∑_{k=1}^{n} F_{2rk}^{3} = ∑_{k=1}^{n} F_{6rk} − 3 ∑_{k=1}^{n} F_{2rk} ,

so that,

\begin{aligned}5L_{3r} ∑_{k=1}^{n} F_{2rk}^{3} & = L_{3r} ∑_{k=1}^{n} F_{6rk} − 3\frac{L_{3r}}{L_{r}} L_{r} ∑_{k=1}^{n} F_{2rk} \\ & = L_{3r} ∑_{k=1}^{n} F_{6rk} − 3(L_{r}^{2} + 3)L_{r} ∑_{k=1}^{n} F_{2rk} . \end{aligned}  (27)

• If n is even, then, by Lemma 2, identity (27) can be written as

5L_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = F_{3rn}L_{3rn+3r} − 3(L_{r}^{2} + 3)F_{rn}L_{rn+r} ,

so that

\frac{5L_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{F_{rn}L_{rn+r}} = \frac{F_{3rn}L_{3rn+3r}}{F_{rn}L_{rn+r}} − 3(L_{r}^{2} + 3)
= 5L_{rn}F_{rn+r}F_{2rn+r} + L_{2r} − 1 − 3L_{r}^{2} − 9, by Lemma 7
= 5L_{rn}F_{rn+r}F_{2rn+r} − 10F_{r}^{2}, by (23) and (24) .

• If n is odd, then, by Lemma 2, we have

5L_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = L_{3rn}F_{3rn+3r} − 3(L_{r}^{2} + 3)L_{rn}F_{rn+r} ,

so that

\frac{5L_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{L_{rn}F_{rn+r}} = \frac{L_{3rn}F_{3rn+3r}}{L_{rn}F_{rn+r}} − 3(L_{r}^{2} + 3)
= 5F_{rn}L_{rn+r}F_{2rn+r} + L_{2r} − 1 − 3L_{r}^{2} − 9, by Lemma 7
= 5F_{rn}L_{rn+r}F_{2rn+r} − 10F_{r}^{2}, by (23) and (24) .

Theorem 13. If r and n are integers such that r is even, then

5F_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = F_{rn}F_{rn+r}(L_{rn}L_{rn+r}L_{2rn+r} − 2L_{r}^{2}) .

Proof.

5F_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = F_{3r} ∑_{k=1}^{n} F_{6rk} − 3\frac{F_{3r}}{F_{r}} F_{r} ∑_{k=1}^{n} F_{2rk}

= F_{3rn}F_{3rn+3r} − 3(5F_{r}^{2} + 3)F_{rn}F_{rn+r}, by Lemma 2 and identity (21) ,

so that

\frac{5F_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{F_{rn}F_{rn+r}} = \frac{F_{3rn}F_{3rn+3r}}{F_{rn}F_{rn+r}} − 3(5F_{r}^{2} + 3)

= L_{rn}L_{rn+r}L_{2rn+r} + L_{2r} − 1 − 15F_{r}^{2} − 9

(by Lemma 4 and identity (21)) , = L_{rn}L_{rn+r}L_{2rn+r} − 2L_{r}^{2}, by (23), (24) and (26) .

Theorem 14. If r and n are integers such that r is odd, then

L_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = \begin{cases}F_{rn}^{2}L_{rn+r}^{2}(L_{rn}F_{rn+r} + F_{r}), & if n is even; \\ L_{rn}^{2}F_{rn+r}^{2}(F_{rn}L_{rn+r} + F_{r}), & if n is odd. \end{cases}

Proof. • If n is even, then from Lemma 2 and identity (27) we have

\frac{5L_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{F_{rn}L_{rn+r}} = \frac{F_{3rn}L_{3rn+3r}}{F_{rn}L_{rn+r}} − 3(L_{r}^{2} + 3)

= 5F_{2rn+r}^{2} + 5F_{rn+r}^{2} + 5F_{rn}^{2} + 5F_{r}^{2} − 3 − 3L_{r}^{2} − 9, by Lemma 11 = 5F_{2rn+r}^{2} + 5F_{rn+r}^{2} + 5F_{rn}^{2} − 10F_{r}^{2} by identity (23) ,

so that

\begin{aligned}\frac{L_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{F_{rn}L_{rn+r}} & = F_{2rn+r}^{2} + F_{rn+r}^{2} + F_{rn}^{2} − 2F_{r}^{2} \\ & = (F_{2rn+r}^{2} − F_{r}^{2}) + (F_{rn+r}^{2} + F_{rn}^{2}) − F_{r}^{2} . \end{aligned}

Using the following identity, derived by Howard [4],

F_{u}^{2} + (−1)^{u+v−1}F_{v}^{2} = F_{u−v}F_{u+v} ,  (28)

we have

\frac{L_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{F_{rn}L_{rn+r}} = F_{2rn}F_{2rn+2r} + F_{r}F_{2rn+r} − F_{r}^{2}

= F_{2rn}F_{2rn+2r} + F_{r}(F_{2rn+r} − F_{r}) = F_{2rn}F_{2rn+2r} + F_{r}F_{rn}L_{rn+r}, by identity (15) = F_{rn}L_{rn+r}L_{rn}F_{rn+r} + F_{r}F_{rn}L_{rn+r} = F_{rn}L_{rn+r}(L_{rn}F_{rn+r} + F_{r}) .

• If n is odd, then from Lemma 2 and identity (27) we have

\frac{5L_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{L_{rn}F_{rn+r}} = \frac{L_{3rn}F_{3rn+3r}}{L_{rn}F_{rn+r}} − 3(L_{r}^{2} + 3)

= 5F_{2rn+r}^{2} + 5F_{rn+r}^{2} + 5F_{rn}^{2} + 5F_{r}^{2} − 3 − 3L_{r}^{2} − 9, by Lemma 10 = 5F_{2rn+r}^{2} + 5F_{rn+r}^{2} + 5F_{rn}^{2} − 10F_{r}^{2} by identity (23) ,

so that

\begin{aligned}\frac{L_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{L_{rn}F_{rn+r}} & = F_{2rn+r}^{2} + F_{rn+r}^{2} + F_{rn}^{2} − 2F_{r}^{2} \\ & = (F_{2rn+r}^{2} − F_{r}^{2}) + (F_{rn+r}^{2} + F_{rn}^{2}) − F_{r}^{2} . \end{aligned}

Using identity (28), we have

\frac{L_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{L_{rn}F_{rn+r}} = F_{2rn}F_{2rn+2r} + F_{r}F_{2rn+r} − F_{r}^{2}

= F_{2rn}F_{2rn+2r} + F_{r}(F_{2rn+r} − F_{r}) = F_{2rn}F_{2rn+2r} + F_{r}L_{rn}F_{rn+r}, by identity (16) = F_{rn}L_{rn+r}L_{rn}F_{rn+r} + F_{r}L_{rn}F_{rn+r} = L_{rn}F_{rn+r}(F_{rn}L_{rn+r} + F_{r}) .

Theorem 15. If r and n are integers such that r is even, then

F_{3r} ∑_{k=1}^{n} F_{2rk}^{3} = F_{rn}^{2}F_{rn+r}^{2}(L_{rn}L_{rn+r} + L_{r}) .  (29)

Proof.

\begin{aligned}5F_{3r} ∑_{k=1}^{n} F_{2rk}^{3} & = F_{3r} ∑_{k=1}^{n} F_{6rk} − 3\frac{F_{3r}}{F_{r}} F_{r} ∑_{k=1}^{n} F_{2rk} \\ & = F_{3rn}F_{3rn+3r} − 3(5F_{r}^{2} + 3)F_{rn}F_{rn+r} , \end{aligned}

so that

\frac{5F_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{F_{rn}F_{rn+r}} = \frac{F_{3rn}F_{3rn+3r}}{F_{rn}F_{rn+r}} − 3(5F_{r}^{2} + 3)
= L_{2rn+r}^{2} + L_{rn+r}^{2} + L_{rn}^{2} + L_{r}^{2} − 7 − 15F_{r}^{2} − 9, by Lemma 8
= L_{2rn+r}^{2} + L_{rn+r}^{2} − 2L_{r}^{2} + 5F_{rn}^{2}, by (23)
= (L_{2rn+r}^{2} − L_{r}^{2}) + (L_{rn+r}^{2} − L_{r}^{2}) + 5F_{rn}^{2} .

Using the identity (derived by Howard [4])

L_{u}^{2} + (−1)^{u+v−1}L_{v}^{2} = 5F_{u−v}F_{u+v} ,  (30)

we see that

L_{2rn+r}^{2} − L_{r}^{2} = 5F_{2rn}F_{2rn+2r} = 5F_{rn}F_{rn+r}L_{rn}L_{rn+r}  (31)

and

L_{rn+r}^{2} − L_{r}^{2} = 5F_{rn}F_{rn+2r} .  (32)

Thus,

\frac{F_{3r} ∑_{k=1}^{n} F_{2rk}^{3}}{F_{rn}F_{rn+r}} = F_{rn}F_{rn+r}L_{rn}L_{rn+r} + F_{rn}F_{rn+2r} + F_{rn}^{2}
= F_{rn}F_{rn+r}L_{rn}L_{rn+r} + F_{rn}(F_{rn} + F_{rn+2r})
= F_{rn}F_{rn+r}L_{rn}L_{rn+r} + F_{rn}F_{rn+r}L_{r}, by identity (16)
= F_{rn}F_{rn+r}(L_{rn}L_{rn+r} + L_{r}) .

3.2 Sums of cubes of Lucas numbers

Theorem 16. If r and n are integers such that r is odd, then

L_{3r} ∑_{k=1}^{n} L_{2rk}^{3} = \begin{cases}5F_{rn}F_{rn+r}(L_{rn}L_{rn+r}L_{2rn+r} + 4(L_{2r} + 1)), & if n is even; \\ L_{rn}L_{rn+r}(5F_{rn}F_{rn+r}L_{2rn+r} + 4(L_{2r} + 1)), & if n is odd. \end{cases}

Proof. Using identity (25) with u = 2rk, we have

∑_{k=1}^{n} L_{2rk}^{3} = ∑_{k=1}^{n} L_{6rk} + 3 ∑_{k=1}^{n} L_{2rk} ,

so that

\begin{aligned}L_{3r} ∑_{k=1}^{n} L_{2rk}^{3} & = L_{3r} ∑_{k=1}^{n} L_{6rk} + 3\frac{L_{3r}}{L_{r}} L_{r} ∑_{k=1}^{n} L_{2rk} \\ & = L_{3r} ∑_{k=1}^{n} L_{6rk} + 3(L_{r}^{2} + 3)L_{r} ∑_{k=1}^{n} L_{2rk}, by (25) . \end{aligned}

• If n is even, then by Lemma 3 we have

L_{3r} ∑_{k=1}^{n} L_{2rk}^{3} = 5F_{3rn}F_{3rn+3r} + 3(L_{r}^{2} + 3)5F_{rn}F_{rn+r} ,  (33)

so that

\frac{L_{3r} ∑_{k=1}^{n} L_{2rk}^{3}}{5F_{rn}F_{rn+r}} = \frac{F_{3rn}F_{3rn+3r}}{F_{rn}F_{rn+r}} + 3(L_{r}^{2} + 3)
= L_{rn}L_{rn+r}L_{2rn+r} + L_{2r} + 1 + 3L_{r}^{2} + 9, by Lemma 4
= L_{rn}L_{rn+r}L_{2rn+r} + 4(L_{2r} + 1), by (24) .

• If n is odd, then by Lemma 3 we have

L_{3r} ∑_{k=1}^{n} L_{2rk}^{3} = L_{3rn}L_{3rn+3r} + 3(L_{r}^{2} + 3)L_{rn}L_{rn+r} ,  (34)

so that

\frac{L_{3r} ∑_{k=1}^{n} L_{2rk}^{3}}{L_{rn}L_{rn+r}} = \frac{L_{3rn}L_{3rn+3r}}{L_{rn}L_{rn+r}} + 3(L_{r}^{2} + 3)
= 5F_{rn}F_{rn+r}L_{2rn+r} + L_{2r} + 1 + 3L_{r}^{2} + 9, by Lemma 5
= 5F_{rn}F_{rn+r}L_{2rn+r} + 4(L_{2r} + 1), by (24) .

F_{3r} ∑_{k=1}^{n} L_{2rk}^{3} = F_{rn}L_{rn+r}(5L_{rn}F_{rn+r}F_{2rn+r} + 4(L_{2r} + 1)) .

Proof.

F_{3r} ∑_{k=1}^{n} L_{2rk}^{3} = F_{3r} ∑_{k=1}^{n} L_{6rk} + 3\frac{F_{3r}}{F_{r}} F_{r} ∑_{k=1}^{n} L_{2rk}

= F_{3r} ∑_{k=1}^{n} L_{6rk} + 3(5F_{r}^{2} + 3)F_{r} ∑_{k=1}^{n} L_{2rk}, by identity (21)

= F_{3rn}L_{3rn+3r} + 3(5F_{r}^{2} + 3)F_{rn}L_{rn+r}, by Lemma 3 .

Thus,

\frac{F_{3r} ∑_{k=1}^{n} L_{2rk}^{3}}{F_{rn}L_{rn+r}} = \frac{F_{3rn}L_{3rn+3r}}{F_{rn}L_{rn+r}} + 3(5F_{r}^{2} + 3)
= 5L_{rn}F_{rn+r}F_{2rn+r} + L_{2r} + 1 + 15F_{r}^{2} + 9, by Lemma 7
= 5L_{rn}F_{rn+r}F_{2rn+r} + 4(L_{2r} + 1), by (24) and (26) .

4 Acknowledgment

Thanks are due to the referee for a careful reading and for pointing out a couple of typographical errors.

References

[1] K. Adegoke, Generalizations of the reciprocal Fibonacci-Lucas sums of Brousseau, J. Integer Seq. 21 (2018), Article 18.1.6.

[2] S. Clary and P. D. Hemenway, On sums of cubes of Fibonacci numbers, in G. E. Bergum, A. N. Philippou and A. F. Horadam, eds., Applications of Fibonacci Numbers, Kluwer Academic Publishers, 1993, pp. 123–136.

[3] L. A. G. Dresel, Transformations of Fibonacci-Lucas identities, in G. E. Bergum, A. N. Philippou and A. F. Horadam, eds., Applications of Fibonacci Numbers, Kluwer Academic Publishers, 1993, pp. 169–184.

[4] F. T. Howard, The sum of the squares of two generalized Fibonacci numbers, Fibonacci Quart. 41 (2003), 80–84.

[5] R. S. Melham, Alternating sums of fourth powers of Fibonacci and Lucas numbers, Fibonacci Quart. 38 (2000), 254–259.

2010 Mathematics Subject Classification: Primary 11B39; Secondary 11B37. Keywords: sum of cubes, Fibonacci number, Lucas number, telescoping summation.

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

Received February 11 2018; revised versions received March 8 2018; July 1 2018; July 3 2018. Published in Journal of Integer Sequences, August 22 2018.

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