On quaternions with generalized Fibonacci and Lucas number components

Abstract

In this paper, we give the exponential generating functions for the generalized Fibonacci and generalized Lucas quaternions, respectively. Moreover, we give some new formulas for binomial sums of these quaternions by using their Binet forms.

1 Introduction

The well-known Fibonacci and Lucas sequences are defined by the following recurrence relations: for $n\geq0$,

$$ F_{n+2}=F_{n+1}+F_{n} $$

and

$$ L_{n+2}=L_{n+1}+L_{n}, $$

where $F_{0}=0$, $F_{1}=1$, $L_{0}=2$, and $L_{1}=1$, respectively. Here, $F_{n}$ is the nth Fibonacci number and $L_{n}$ is the nth Lucas number.

The Binet formulas for the Fibonacci and Lucas sequences are given by

$$ F_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta} $$

and

$$ L_{n}=\alpha^{n}+\beta^{n}, $$

where

$$ \alpha=\frac{1+\sqrt{5}}{2}\quad \text{and}\quad \beta=\frac{1-\sqrt{5}}{2}. $$

The numerous relations connecting Fibonacci and Lucas numbers have been given by Vajda [1].

Carlitz, Ferns, Layman and Hoggatt studied many properties of binomial sums of these numbers (see [2–5]).

The generalized Fibonacci and Lucas sequences, $\{U_{n}\}$ and $\{ V_{n}\}$, are defined by the following recurrence relations: for $n\geq0$ and any nonzero integer p,

$$ U_{n+2}=pU_{n+1}+U_{n} $$

and

$$ V_{n+2}=pV_{n+1}+V_{n}, $$

where $U_{0}=0$, $U_{1}=1$, $V_{0}=2$, and $V_{1}=p$, respectively. If we take $p=1$, then $U_{n}=F_{n}$ (nth Fibonacci number) and $V_{n}=L_{n}$ (nth Lucas number).

Let λ and μ be the roots of the characteristic equation $x^{2}-px-1=0$. Then the Binet formulas for the sequences $\{U_{n}\}$ and $\{V_{n}\}$ are given by

$$ U_{n}=\frac{\lambda^{n}-\mu^{n}}{\lambda-\mu} $$

and

$$ V_{n}=\lambda^{n}+\mu^{n}, $$

where $\lambda= ( p+\sqrt{p^{2}+4} ) /2$ and $\mu= ( p-\sqrt{p^{2}+4} ) /2$.

From these equalities, by putting $\Delta=p^{2}+4$, we see that $1+\lambda ^{2}=\lambda\sqrt{\Delta}$ and $1+\mu^{2}=-\mu\sqrt{\Delta}$.

The generating function and the exponential generating function of a sequence $\{ a_{n} \}$ are defined by

$$ a(x)=\sum_{n=0}^{\infty}a_{n}x^{n} $$

and

$$ \hat{a}(x)=\sum_{n=0}^{\infty}a_{n} \frac{x^{n}}{n!}, $$

respectively.

Now, we would like to explain how the generating function of a given sequence $\{ a_{n} \}$ is derived: Firstly, we write down a recurrence relation which is a single equation expressing $a_{n}$ in terms of other elements of the sequence. This equation should be valid for all integers n, assuming that $a_{-1}=a_{-2}=\cdots=0$. Then we multiply both sides of the equation by $x^{n}$ and sum over all n. This gives, on the left, the sum $\sum_{n=0}^{\infty}a_{n}x^{n}$, which is the generating function $a(x)$. The right-hand side should be manipulated so that it becomes some other expression involving $a(x)$. Lastly, if we solve the resulting equation, we get a closed form for $a(x)$. For more details and properties related to the generating functions and special functions, we refer to [6–12].

A quaternion p with real components $a_{0}$, $a_{1}$, $a_{2}$, $a_{3}$, and basis 1, i, j, k is an element of the form

$$ p=a_{0}\mathbf{1}+a_{1}\mathbf {i}+a_{2} \mathbf {j}+a_{3}\mathbf {k}= ( a_{0},a_{1},a_{2},a_{3} ) \quad (a_{0}\mathbf{1}=a_{0}), $$

where

$$\begin{aligned}& \mathbf {i}^{2} = \mathbf {j}^{2}= \mathbf {k}^{2}=-\mathbf{1}, \\& \mathbf {ij} = \mathbf {k}=-\mathbf {ji},\qquad \mathbf {jk}= \mathbf {i}=-\mathbf {kj},\qquad \mathbf {ki}=\mathbf {j}=- \mathbf {ik}. \end{aligned}$$

Throughout this paper, we work in the real division quaternion algebra without specific references being given.

The nth Fibonacci and the nth Lucas quaternions were defined by Horadam in [13] as

$$ Q_{n}=F_{n}+F_{n+1}\mathbf {i}+F_{n+2} \mathbf {j}+F_{n+3}\mathbf {k} $$

and

$$ K_{n}=L_{n}+L_{n+1}\mathbf {i}+L_{n+2} \mathbf {j}+L_{n+3}\mathbf {k}, $$

respectively, where $F_{n}$ and $L_{n}$ are the nth Fibonacci number and the nth Lucas number.

Similar to Horadam, the nth generalized Fibonacci quaternion and nth generalized Lucas quaternion are defined by

$$ \mathcal{Q}_{n}=U_{n}+U_{n+1}\mathbf {i}+U_{n+2} \mathbf {j}+U_{n+3}\mathbf {k} $$

and

$$ \mathcal{K}_{n}=V_{n}+V_{n+1}\mathbf {i}+V_{n+2} \mathbf {j}+V_{n+3}\mathbf {k}, $$

respectively [14]. Here, $U_{n}$ and $V_{n}$ are the nth generalized Fibonacci number and the nth generalized Lucas number.

For $n\geq0$, the following recurrence relations hold:

$$ \mathcal{Q}_{n+2}=p\mathcal{Q}_{n+1}+\mathcal{Q}_{n} $$

and

$$ \mathcal{K}_{n+2}=p\mathcal{K}_{n+1}+\mathcal{K}_{n}, $$

where

$$\begin{aligned}& \mathcal{Q}_{0}= \bigl( 0,1,p,p^{2}+1 \bigr) ,\qquad \mathcal{Q}_{1}= \bigl( 1,p,p^{2}+1,p^{3}+2p \bigr) , \\& \mathcal{K}_{0}= \bigl( 2,p,p^{2}+2,p^{3}+3p \bigr) ,\qquad \mathcal{K}_{1}= \bigl( p,p^{2}+2,p^{3}+3p,p^{4}+4p^{2}+2 \bigr) . \end{aligned}$$

In [15], Iakin gave the Binet formulas for the generalized Fibonacci and generalized Lucas quaternions as follows:

$$ \mathcal{Q}_{n}=\frac{\underline{\lambda}\lambda^{n}-\underline{\mu }\mu ^{n}}{\lambda-\mu} $$

and

$$ \mathcal{K}_{n}=\underline{\lambda}\lambda^{n}+\underline{ \mu}\mu ^{n}, $$

where $\underline{\lambda}=1+\lambda \mathbf {i}+\lambda ^{2}\mathbf {j}+\lambda^{3}\mathbf {k}$ and $\underline{\mu}=1+\mu \mathbf {i}+\mu^{2}\mathbf {j}+\mu^{3}\mathbf {k}$.

Iyer [16] gave some relations connecting the Fibonacci and Lucas quaternions. In [17], Swamy derived the relations of generalized Fibonacci quaternions. Iakin [18, 19] introduced the concept of a higher order quaternion and generalized quaternions with quaternion components. Horadam [20] studied the quaternion recurrence relations. In [21], Halici derived generating functions and many identities for Fibonacci and Lucas quaternions. In [22], Flaut and Shpakivskyi investigated some properties of generalized Fibonacci quaternions and Fibonacci-Narayana quaternions. Very recently, Ramirez [14] has obtained some combinatorial properties of the k-Fibonacci and the k-Lucas quaternions.

Inspired by these results, in the present paper we give the exponential generating functions for the generalized Fibonacci and generalized Lucas quaternions. Moreover, we derive some new formulas for binomial sums of these quaternions by using their Binet forms.

2 Some properties of generalized Fibonacci and Lucas quaternions

Theorem 2.1

([14])

The generating functions for the generalized Fibonacci and generalized Lucas quaternions are

$$ F(t)=\frac{\mathcal{Q}_{0}+(\mathcal{Q}_{1}-p\mathcal{Q}_{0})t}{1-pt-t^{2}} $$

and

$$ L(t)=\frac{\mathcal{K}_{0}+(\mathcal{K}_{1}-p\mathcal {K}_{0})t}{1-pt-t^{2}}, $$

respectively.

Proof

Let $F(t)=\sum_{n=0}^{\infty}\mathcal{Q}_{n}t^{n}$ and $L(t)=\sum_{n=0}^{\infty}\mathcal{K}_{n}t^{n}$. Then we get the following equation:

$$\begin{aligned}& \sum_{n=0}^{\infty}\mathcal{Q}_{n}t^{n}-pt \sum_{n=0}^{\infty}\mathcal {Q}_{n}t^{n}-t^{2} \sum_{n=0}^{\infty}\mathcal{Q}_{n}t^{n} \\& \quad =\mathcal {Q}_{0}+(\mathcal{Q}_{1}-p \mathcal{Q}_{0})t +\sum_{n=2}^{\infty} ( \mathcal{Q}_{n}-p\mathcal {Q}_{n-1}-\mathcal{Q}_{n-2} ) t^{n}. \end{aligned}$$

Since, for each $n\geq2$, the coefficient of $t^{n}$ is zero in the right-hand side of this equation, we obtain

$$ F(t)=\frac{\mathcal{Q}_{0}+(\mathcal{Q}_{1}-p\mathcal {Q}_{0})t}{1-pt-t^{2}}. $$

Similarly, we obtain

$$ L(t)=\frac{\mathcal{K}_{0}+(\mathcal{K}_{1}-p\mathcal {K}_{0})t}{1-pt-t^{2}}. $$

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Theorem 2.2

The exponential generating functions for the generalized Fibonacci and generalized Lucas quaternions are

$$ \sum_{n=0}^{\infty}\frac{\mathcal{Q}_{n}}{n!}t^{n}= \frac{\underline{\lambda}e^{\lambda t}-\underline{\mu}e^{\mu t}}{\lambda-\mu} $$

and

$$ \sum_{n=0}^{\infty}\frac{\mathcal{K}_{n}}{n!}t^{n}= \underline{\lambda}e^{\lambda t}+\underline{\mu}e^{\mu t}. $$

Proof

By the Binet formula for the generalized Fibonacci quaternions, we get

$$\begin{aligned} \sum_{n=0}^{\infty}\frac{\mathcal{Q}_{n}}{n!}t^{n} =&\sum_{n=0}^{\infty} \biggl( \frac{\underline{\lambda}\lambda^{n}-\underline{\mu}\mu^{n}}{\lambda-\mu} \biggr) \frac{t^{n}}{n!} \\ =&\frac{\underline{\lambda}}{\lambda-\mu}\sum_{n=0}^{\infty}\frac{ ( \lambda t ) ^{n}}{n!}-\frac{\underline{\mu}}{\lambda -\mu }\sum_{n=0}^{\infty} \frac{ ( \mu t ) ^{n}}{n!} \\ =&\frac{\underline{\lambda}e^{\lambda t}-\underline{\mu}e^{\mu t}}{\lambda-\mu}. \end{aligned}$$

Similarly, by the Binet formula for the generalized Lucas quaternions we obtain

$$\begin{aligned} \sum_{n=0}^{\infty}\frac{\mathcal{K}_{n}}{n!}t^{n} =&\sum_{n=0}^{\infty} \bigl( \underline{\lambda} \lambda^{n}+\underline{\mu}\mu^{n} \bigr) \frac{t^{n}}{n!} \\ =&\underline{\lambda}\sum_{n=0}^{\infty} \frac{ ( \lambda t ) ^{n}}{n!}+\underline{\mu}\sum_{n=0}^{\infty} \frac{ ( \mu t ) ^{n}}{n!} \\ =&\underline{\lambda}e^{\lambda t}+\underline{\mu}e^{\mu t}. \end{aligned}$$

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Now, we give the following theorems in which the first formulas are proved, and the remaining formulas can be obtained similarly.

Theorem 2.3

For $n,k\geq0$, we have

$$\begin{aligned}& \sum_{i=0}^{n}\binom{n}{i} \mathcal{Q}_{2i+k}=\left \{ \begin{array}{l@{\quad}l} \Delta^{\frac{n}{2}}\mathcal{Q}_{n+k}&\textit{if }n\textit{ is even}, \\ \Delta^{\frac{n-1}{2}}\mathcal{K}_{n+k}&\textit{if }n\textit{ is odd},\end{array} \right . \\& \sum_{i=0}^{n}\binom{n}{i} \mathcal{K}_{2i+k}=\left \{ \begin{array}{l@{\quad}l} \Delta^{\frac{n}{2}}\mathcal{K}_{n+k}&\textit{if }n\textit{ is even}, \\ \Delta^{\frac{n+1}{2}}\mathcal{Q}_{n+k}&\textit{if }n\textit{ is odd},\end{array} \right . \\& \sum_{i=0}^{n}\binom{n}{i}(-1)^{i} \mathcal{Q}_{2i+k}=(-p)^{n}\mathcal{Q}_{n+k}, \\& \sum_{i=0}^{n}\binom{n}{i}(-1)^{i} \mathcal{K}_{2i+k}=(-p)^{n}\mathcal{K}_{n+k}. \end{aligned}$$

Proof

If we use the Binet formula for the generalized Fibonacci quaternions, we get

$$\begin{aligned} \sum_{i=0}^{n}\binom{n}{i} \mathcal{Q}_{2i+k} =&\sum_{i=0}^{n} \binom{n}{i} \biggl( \frac{\underline{\lambda}\lambda^{2i+k}-\underline{\mu}\mu^{2i+k}}{\lambda-\mu} \biggr) \\ =&\frac{\underline{\lambda}\lambda^{k}}{\lambda-\mu}\sum_{i=0}^{n} \binom{n}{i}\lambda^{2i}-\frac{\underline{\mu}\mu^{k}}{\lambda-\mu}\sum _{i=0}^{n}\binom{n}{i}\mu^{2i} \\ =&\frac{\underline{\lambda}\lambda^{k}}{\lambda-\mu} \bigl( 1+\lambda ^{2} \bigr) ^{n}- \frac{\underline{\mu}\mu^{k}}{\lambda-\mu} \bigl( 1+\mu ^{2} \bigr) ^{n} \\ =&\frac{\underline{\lambda}\lambda^{k}}{\lambda-\mu} ( \lambda \sqrt{\Delta} ) ^{n}- \frac{\underline{\mu}\mu^{k}}{\lambda-\mu } ( -\mu\sqrt{\Delta} ) ^{n} \\ =&\left \{ \begin{array}{l@{\quad}l} \Delta^{\frac{n}{2}}\mathcal{Q}_{n+k} &\text{if }n\text{ is even}, \\ \Delta^{\frac{n-1}{2}}\mathcal{K}_{n+k}&\text{if }n\text{ is odd}.\end{array} \right . \end{aligned}$$

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Theorem 2.4

([14])

For $n\geq0$, we have

$$\begin{aligned}& \sum_{i=0}^{n}\binom{n}{i}p^{i} \mathcal{Q}_{i}=\mathcal{Q}_{2n}, \\& \sum_{i=0}^{n}\binom{n}{i}p^{i} \mathcal{K}_{i}=\mathcal{K}_{2n}. \end{aligned}$$

Proof

By the Binet formula for the generalized Fibonacci quaternions, we have

$$\begin{aligned} \sum_{i=0}^{n}\binom{n}{i}p^{i} \mathcal{Q}_{i} =&\sum_{i=0}^{n} \binom{n}{i}p^{i} \biggl( \frac{\underline{\lambda}\lambda^{i}-\underline{\mu}\mu^{i}}{\lambda-\mu} \biggr) \\ =&\frac{\underline{\lambda}}{\lambda-\mu}\sum_{i=0}^{n} \binom{n}{i} ( p\lambda ) ^{i}-\frac{\underline{\mu}}{\lambda-\mu}\sum _{i=0}^{n}\binom{n}{i} ( p\mu ) ^{i} \\ =&\frac{\underline{\lambda}}{\lambda-\mu} ( 1+p\lambda ) ^{n}-\frac{\underline{\mu}}{\lambda-\mu} ( 1+p\mu ) ^{n} \\ =&\frac{1}{\lambda-\mu} \bigl( \underline{\lambda}\lambda^{2n}-\underline{\mu}\mu^{2n} \bigr) \\ =&\mathcal{Q}_{2n}. \end{aligned}$$

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Theorem 2.5

For $n\geq0$, we have

$$\begin{aligned}& \sum_{i=0}^{n}\binom{n}{2i} \mathcal{Q}_{4i}=\left \{ \begin{array}{l@{\quad}l} \frac{1}{2} ( \Delta^{\frac{n}{2}}+p^{n} ) \mathcal {Q}_{n}&\textit{if }n\textit{ is even}, \\ \frac{1}{2} ( \Delta^{\frac{n-1}{2}}\mathcal{K}_{n}-p^{n}\mathcal {Q}_{n} )& \textit{if }n\textit{ is odd},\end{array} \right . \\& \sum_{i=0}^{n}\binom{n}{2i} \mathcal{K}_{4i}=\left \{ \begin{array}{l@{\quad}l} \frac{1}{2} ( \Delta^{\frac{n}{2}}+p^{n} ) \mathcal {K}_{n}&\textit{if }n\textit{ is even}, \\ \frac{1}{2} ( \Delta^{\frac{n+1}{2}}\mathcal{Q}_{n}-p^{n}\mathcal {K}_{n} )& \textit{if }n\textit{ is odd},\end{array} \right . \\& \sum_{i=0}^{n}\binom{n}{2i+1} \mathcal{Q}_{4i+1}=\left \{ \begin{array}{l@{\quad}l} \frac{1}{2} ( \Delta^{\frac{n}{2}}-p^{n} ) \mathcal {Q}_{n}&\textit{if }n\textit{ is even}, \\ \frac{1}{2} ( \Delta^{\frac{n-1}{2}}\mathcal{K}_{n}+p^{n}\mathcal {Q}_{n} )& \textit{if }n\textit{ is odd},\end{array} \right . \\& \sum_{i=0}^{n}\binom{n}{2i+1} \mathcal{K}_{4i+1}=\left \{ \begin{array}{l@{\quad}l} \frac{1}{2} ( \Delta^{\frac{n}{2}}-p^{n} ) \mathcal {K}_{n}&\textit{if }n\textit{ is even}, \\ \frac{1}{2} ( \Delta^{\frac{n+1}{2}}\mathcal{Q}_{n}+p^{n}\mathcal {K}_{n} ) &\textit{if }n\textit{ is odd}.\end{array} \right . \end{aligned}$$

Proof

By Theorem 2.3, we obtain

$$\begin{aligned} \sum_{i=0}^{n}\binom{n}{2i} \mathcal{Q}_{4i} =&\frac{1}{2}\sum _{i=0}^{n}\binom{n}{i} \bigl( 1+(-1)^{i} \bigr) \mathcal{Q}_{2i} \\ =&\frac{1}{2} \Biggl[ \sum_{i=0}^{n} \binom{n}{i}\mathcal{Q}_{2i}+\sum _{i=0}^{n}\binom{n}{i}(-1)^{i} \mathcal{Q}_{2i} \Biggr] \\ =&\left \{ \begin{array}{l@{\quad}l} \frac{1}{2} ( \Delta^{\frac{n}{2}}+p^{n} ) \mathcal {Q}_{n}&\text{if }n\text{ is even}, \\ \frac{1}{2} ( \Delta^{\frac{n-1}{2}}\mathcal{K}_{n}-p^{n}\mathcal {Q}_{n} ) &\text{if }n\text{ is odd}.\end{array} \right . \end{aligned}$$

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Theorem 2.6

For $n\geq0$, we have

$$\begin{aligned}& \sum_{i=0}^{n}\binom{n}{i} ( \mathcal{Q}_{i} ) ^{2}=\left \{ \begin{array}{l@{\quad}l} \Delta^{\frac{n-2}{2}} ( 2\mathcal{K}_{n}-(V_{5}+p)V_{n+1}-(V_{4}+V_{0})V_{n} ) & \textit{if }n \textit{ is even}, \\ \Delta^{\frac{n-1}{2}} ( 2\mathcal{Q}_{n}-(V_{5}+p)U_{n+1}-(V_{4}+V_{0})U_{n} )& \textit{if }n \textit{ is odd},\end{array} \right . \\& \sum_{i=0}^{n}\binom{n}{i} ( \mathcal{K}_{i} ) ^{2}=\left \{ \begin{array}{l@{\quad}l} \Delta^{\frac{n}{2}} ( 2\mathcal{K}_{n}-(V_{5}+p)V_{n+1}-(V_{4}+V_{0})V_{n} )& \textit{if }n\textit{ is even}, \\ \Delta^{\frac{n+1}{2}} ( 2\mathcal{Q}_{n}-(V_{5}+p)U_{n+1}-(V_{4}+V_{0})U_{n} )& \textit{if }n\textit{ is odd}.\end{array} \right . \end{aligned}$$

Proof

From the quaternion multiplication, we have

$$\begin{aligned}& ( \underline{\lambda} ) ^{2}=2\underline{\lambda}-\bigl( ( V_{5}+p ) \lambda+V_{4}+V_{0}\bigr), \\& ( \underline{\mu} ) ^{2}=2\underline{\mu}-\bigl( ( V_{5}+p ) \mu+V_{4}+V_{0}\bigr). \end{aligned}$$

If we consider the Binet formula for the generalized Fibonacci quaternions, we have

$$\begin{aligned} \sum_{i=0}^{n}\binom{n}{i} ( \mathcal{Q}_{i} ) ^{2} =&\sum_{i=0}^{n} \binom{n}{i} \biggl( \frac{\underline{\lambda}\lambda^{i}-\underline{\mu}\mu^{i}}{\lambda-\mu} \biggr) ^{2} \\ =&\frac{1}{\Delta}\sum_{i=0}^{n} \binom{n}{i} \bigl[ ( \underline{\lambda} ) ^{2} \lambda^{2i}+ ( \underline{\mu } ) ^{2}\mu^{2i}-2 \underline{\lambda}\underline{\mu} ( \lambda\mu ) ^{i} \bigr] \\ =&\frac{1}{\Delta}\sum_{i=0}^{n} \binom{n}{i} \bigl[ ( \underline{\lambda} ) ^{2} \lambda^{2i}+ ( \underline{\mu } ) ^{2}\mu^{2i} \bigr] -\frac{2\underline{\lambda}\underline{\mu }}{\Delta}\sum_{i=0}^{n} \binom{n}{i} ( -1 ) ^{i} \\ =&\frac{ ( \underline{\lambda} ) ^{2}}{\Delta}\sum_{i=0}^{n} \binom{n}{i}\lambda^{2i}+\frac{ ( \underline{\mu } ) ^{2}}{\Delta}\sum _{i=0}^{n}\binom{n}{i}\mu^{2i} \\ =&\frac{ ( \underline{\lambda} ) ^{2}}{\Delta} \bigl( 1+\lambda ^{2} \bigr) ^{n}+ \frac{ ( \underline{\mu} ) ^{2}}{\Delta } \bigl( 1+\mu^{2} \bigr) ^{n} \\ =&\frac{ ( \underline{\lambda} ) ^{2}}{\Delta} ( \lambda \sqrt{\Delta} ) ^{n}+ \frac{ ( \underline{\mu} ) ^{2}}{\Delta } ( -\mu\sqrt{\Delta} ) ^{n} \\ =&\Delta^{\frac{n-2}{2}} \bigl( ( \underline{\lambda} ) ^{2} \lambda^{n}+ ( \underline{\mu} ) ^{2} ( -\mu ) ^{n} \bigr) \\ =&\left \{ \begin{array}{l@{\quad}l} \Delta^{\frac{n-2}{2}} ( 2\mathcal{K}_{n}-(V_{5}+p)V_{n+1}-(V_{4}+V_{0})V_{n} ) &\text{if }n\text{ is even}, \\ \Delta^{\frac{n-1}{2}} ( 2\mathcal{Q}_{n}-(V_{5}+p)U_{n+1}-(V_{4}+V_{0})U_{n} )& \text{if }n\text{ is odd}.\end{array} \right . \end{aligned}$$

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Note that, by taking $p=1$ in Theorem 2.6, we obtain

$$ \sum_{i=0}^{n}\binom{n}{i} ( Q_{i} ) ^{2}=\left \{ \begin{array}{l@{\quad}l} 5^{\frac{n-2}{2}} ( 2K_{n}-L_{n}-4L_{n+4} ) &\text{if } n \text{ is even}, \\ 5^{\frac{n-1}{2}} ( 2Q_{n}-F_{n}-4F_{n+4} )& \text{if } n \text{ is odd} \end{array} \right . $$

and

$$ \sum_{i=0}^{n}\binom{n}{i} ( K_{i} ) ^{2}=\left \{ \begin{array}{l@{\quad}l} 5^{\frac{n}{2}} ( 2K_{n}-L_{n}-4L_{n+4} )& \text{if }n \text{ is even}, \\ 5^{\frac{n+1}{2}} ( 2Q_{n}-F_{n}-4F_{n+4} ) &\text{if } n \text{ is odd}, \end{array} \right . $$

where $Q_{i}$ and $K_{i}$ are the ith Fibonacci quaternion and the ith Lucas quaternion, respectively.