Fibonacci polynomials have been generalized mainly by two ways: by maintaining the recurrence relation and varying the initial conditions and by varying the recurrence relation and maintaining the initial conditions. In this paper, both the recurrence relation and initial conditions of generalized Fibonacci polynomials are varied and defined by recurrence relation as \(R_n(x)=axR_{n-1}(x)+bR_{n-2}(x)\) for all \(n\geq2,\) with initial conditions \(R_0(x)=2p\) and \(R_1(x)=px+q\) where \(a\) and \(b\) are positive integers and \(p\) and \(q\) are non-negative integers. Further some fundamental properties of these generalized polynomials such as explicit sum formula, sum of first \(n\) terms, sum of first \(n\) terms with (odd or even) indices and generalized identity are derived by Binet’s formula and generating function only.
Fibonacci polynomials [1] are special cases of Chebyshev polynomials and are defined recursively by
Lucas polynomials [1] are defined by
Pell polynomials [2] are defined by
Generating function of Fibonacci polynomials is given by
Generalized Fibonacci polynomials and some identities [4] are defined by
\[w_n(x)=xw_{n-1}(x)+w_{n-2}(x),\] for \(n\geq2\) with \(w_0(x)=2b\) and \(w_1(x)=a+b,\) where \(a\) and \(b\) are integers.Generalized Fibonacci-Like polynomials [5] are defined by the recurrence relation:
\[m_n(x)=xm_{n-1}(x)+m_{n-2}(x),\] with \(m_0(x)=2s\) and \(m_1(x)=1+s,\) where \(s\) is an integer.Fibonacci-like polynomials [6] are defined by the recurrence relation
\[S_n(x)=xS_{n-1}(x)+S_{n-2}(x),\] \(n\geq2\) with \(S_0(x)=2\) and \(S_1(x)=2x. \)Generalized Fibonacci polynomials [7] are defined by the recurrence relation
\[b_n(x)=xb_{n-1}(x)+b_{n-2}(x),\] \(n\geq2\) with \(b_0(x)=2b\) and \(b_1(x)=s,\) where \(b\) and \(s\) are integer.In this paper, generalized Fibonacci polynomials is studied by varying both the recurrence relation and initial conditions. The properties of these polynomials are derived by means of Binet’s formula and generating function. Few terms of generalized Fibonacci polynomials and characteristic equation of the recurrence relation are presented in Section 2. In Section 3 Binet’s formula is obtained and generating function is also obtained in Section 4. Further some properties of these polynomials are presented in Section 5 and finally in Section 6 conclusion is given.
Few terms of generalized Fibonacci sequence are as follows:
\begin{align*} R_0(x)&=2p,\\ R_1(x)&=px+q,\\ R_2(x)&=apx^2+aqx+2bp,\\ R_3(x)&=a^2px^3+a^2qx^2+2abpx+bpx+bq,\\ R_4(x)&=a^3px^4+a^3qx^3+2a^2bpx^2+abpx^2+abqx+abpx^2+abqx+2b^2p \end{align*} and so on.The characteristic equation for the recurrence relation (6) is
Theorem [Binet’s formula] The \(n^{th}\) term of generalized Fibonacci polynomials is given by
Proof. The characteristics Equation (7) has real and distinct roots. The solution of the recurrence relation (6) is therefore of the form
Setting \(n=0\) and \(n=1\) in (9), we obtain
\[A+B=R_0(x)\] and \[A\alpha(x)+B\beta(x)=R_1(x)\] respectively.Solving these equations simultaneously, we obtain
\[A=\dfrac{R_1(x)- \beta(x)R_0(x)}{\alpha(x)-\beta(x)}\] and \[B=\dfrac{\alpha(x)R_0(x)-R_1(x)}{\alpha(x)-\beta(x)}.\] Substituting for \(A\) and \(B\) in (9), we get \begin{align*} R_n(x)&={\left(\dfrac{R_1(x)-\beta(x)R_0(x)}{\alpha(x)-\beta(x)}\right)}\alpha^n(x)-{\left(\dfrac{R_1(x)-\alpha(x)R_0(x)}{\alpha(x)-\beta(x)}\right)}\beta^n(x)\\ &=\dfrac{1}{\alpha(x)-\beta(x)}\left[R_1(x)(\alpha^n(x)-\beta^n(x))+bR_0(x)(\alpha^{n-1}(x)-\beta^{n-1}(x))\right]. \end{align*} Hence the proof.Remark 1. We have that
Theorem 2. [Generating function] Generating function for generalized Fibonacci polynomials is given by
Proof. Applying power series to the generalized Fibonacci polynomial \(\displaystyle\sum_{n=0}^{\infty}R_n(x)t^n,\) we have \[2p+(px+q)t+(ax^2p+axq+2bp)t^2+\cdots=\displaystyle\sum_{n=0}^{\infty}R_n(x)t^n. \] Now, multiplying the generating series by \((1-axt-bt^2),\) where \((1-axt-bt^2) \neq 0,\) we get \begin{align*} &(1-axt-bt^2)\displaystyle\sum_{n=0}^{\infty}R_n(x)t^n=\displaystyle\sum_{n=0}^{\infty}R_n(x)t^n-ax\displaystyle\sum_{n=0}^{\infty}R_n(x)t^{n+1}-b\displaystyle\sum_{n=0}^{\infty}R_n(x)t^{n+2}\\ &={\left[R_0(x)+R_1(x)t +\displaystyle\sum_{n=2}^{\infty}R_n(x)t^n\right]}-ax{\left[R_0(x)t+\displaystyle\sum_{n=2}^{\infty}R_{n-1}(x)t^n\right]}-b\displaystyle\sum_{n=2}^{\infty}R_{n-2}(x)t^n\\ &=R_0(x)+{\left[R_1(x)-axR_0(x)\right]}t+\displaystyle\sum_{n=2}^{\infty}{\left[R_n(x)-axR_{n-1}(x)-bR_{n-2}(x)\right]}t^n\\ &=2p+{\left[(px+q)-2apx\right]}t+\displaystyle\sum_{n=2}^{\infty}{\left[axR_{n-1}(x)+bR_{n-2}(x)-axR_{n-1}(x)-bR_{n-2}(x)\right]}\\ &=2p+{\left[(px+q)-2apx\right]t}. \end{align*} Therefore \[{\left(1-axt-bt^2\right)}\displaystyle\sum_{n=0}^{\infty}R_n(x)t^n=2p+{\left[(px+q)-2apx\right]t}.\] Hence \[\displaystyle\sum_{n=0}^{\infty}R_n(x)t^n=\dfrac{2p+[(px+q)-2apx]t}{1-axt-bt^2}.\]
Remark 2. If \(a=b=q=1\) and \(p=0\) in (15), we obtain generating functions for Fibonacci polynomials (4).
Proposition 1. [Explicit sum formula] Let \(R_n(x)\) be the \(n^{th}\) generalized Fibonacci polynomials, then
Proof. By generating function (15), we have \begin{align*} \displaystyle\sum_{n=0}^{\infty}R_n(x)t^n&=\dfrac{2p+(px+q-2apx)t}{1-axt-bt^2}\\ &={\left[2p+(px+q-2apx)t\right]}{\left[1-(ax+bt)t\right]^{-1}}\\ &={\left[2p+(px+q-2apx)t\right]}\displaystyle\sum_{n=0}^{\infty}(ax+bt)^nt^n\\ &={\left[2p+(px+q-2apx)t\right]}\displaystyle\sum_{n=0}^{\infty}t^n\displaystyle\sum_{k=0}^n{n\choose k}(ax)^{n-k}(bt)^k\\ &={\left[2p+(px+q-2apx)t\right]}\displaystyle\sum_{n=0}^{\infty}\displaystyle\sum_{k=0}^n \dfrac{n!}{k!(n-k)!}(ax)^{n-k}b^kt^{n+k}. \end{align*} Now replacing \(n\) with \(n+k, \) we get \begin{align*} \displaystyle\sum_{n=0}^{\infty}R_n(x)t^n&={\left[2p+(px+q-2apx)t\right]}\displaystyle\sum_{n=0}^{\infty}\displaystyle\sum_{k=0}^{\infty} \dfrac{(n+k)!}{k!n!}(ax)^n b^k t^{n+2k}\\ &={\left[2p+(px+q-2apx)t\right]}\displaystyle\sum_{n=0}^{\infty}\displaystyle\sum_{n=0}^{\left\lfloor\dfrac{n}{2}\right\rfloor} \dfrac{(n-k)!}{k!(n-2k)!} (ax)^{n-2k} b^k t^n. \end{align*} Thus the sum equals to \(\displaystyle\sum_{n=0}^{\infty}{\left[2p\displaystyle\sum_{k=0}^{\left\lfloor\dfrac{n}{2}\right\rfloor} \dfrac{(n-k)!}{k!(n-2k)!} (ax)^{n-2k} b^k\right]t^n}+\displaystyle\sum_{n=0}^{\infty}{\left[(px+q-2apx)\displaystyle\sum_{k=0}^{\left\lfloor\dfrac{n}{2}\right\rfloor}\dfrac{(n-k)!}{k!(n-2k)!}(ax)^{n-2k} b^k\right]t^{n+1}}.\) Equating the coefficient of \(t^n\) on both sides, we obtain \(R_n(x)=2p\displaystyle\sum_{k=0}^{\left\lfloor\dfrac{n}{2}\right\rfloor}{n-k\choose k}(ax)^{n-2k} b^k +{\left(px+q-2apx\right)}\displaystyle\sum_{k=0}^{\left\lfloor\dfrac{n-1}{2}\right\rfloor} {n-k-1 \choose k}(ax)^{n-2k-1} b^k.\)
Proposition 2. [Sum of first \(n\) terms] The sum of the first \(n\) terms of generalized Fibonacci polynomials is given by \[\displaystyle\sum_{k=0}^{n-1}R_k(x)=\dfrac{R_n(x)+bR_{n-1}(x)-(R_1(x)-axR_0(x))-R_0(x)}{ax+b-1}.\]
Proof. Using Binet’s formula (8), we have \begin{align*} \displaystyle\sum_{k=0}^{n-1}R_k(x)&=\displaystyle\sum_{k=0}^{n-1}{\left(A\alpha^{k}(x)+B\beta^{k}(x)\right)}, \end{align*} where \(A=\dfrac{R_1(x)- \beta(x)R_0(x)}{\alpha(x)-\beta(x)}\) and \(B=\dfrac{\alpha(x)R_0(x)-R_1(x)}{\alpha(x)-\beta(x)}.\) It follows that \begin{align*} \displaystyle\sum_{k=0}^{n-1}R_k(x)&=A\displaystyle\sum_{k=0}^{n-1}\alpha^k(x)+B\displaystyle\sum_{k=0}^{n-1}\beta^k(x)\\ &=\dfrac{A\left(\alpha^n(x)-1\right)}{\alpha(x)-1}+\dfrac{B\left(\beta^n(x)-1\right)}{\beta(x)-1}\\ &={\dfrac{A+B-(A\beta(x)+B\alpha(x))-(A\alpha^{n}(x)+B\beta^{n}(x))}{\alpha(x)\beta(x)-\alpha(x)-\beta(x)+1}} +{\dfrac{\alpha(x)\beta(x)(A\alpha^{n-1}(x) +B\beta^{n-1}(x))}{\alpha(x)\beta(x)-\alpha(x)-\beta(x)+1}}. \end{align*} Since \(\alpha(x)+\beta(x)=ax\) and \(\alpha(x)\beta(x)=-b\) and using (8),(10) and (12), we obtain \[\displaystyle\sum_{k=0}^{n-1}R_k(x)=\dfrac{R_n(x)+bR_{n-1}(x)-(R_1(x)-axR_0(x))-R_0(x)}{ax+b-1}.\]
Proposition 3. [Sum of first \(n\) terms with odd indices] The sum of first \(n\) terms with odd indices of generalized Fibonacci polynomials is given by \[\displaystyle\sum_{k=0}^{n-1}R_{2k+1}(x)=\dfrac{R_{2n+1}(x)-b^2R_{2n-1}(x)+b{\left(R_1(x)-axR_0(x)\right)}-R_1(x)}{a^2x^2-b^2+2b-1}.\]
Proof. Using Binet’s formula (8), we have \begin{align*} \displaystyle\sum_{k=0}^{n-1}R_{2k+1}(x)&=\displaystyle\sum_{k=0}^{n-1}{\left(A\alpha^{2k+1}(x)+B\beta^{2k+1}(x)\right)}\\ &=A\displaystyle\sum_{k=0}^{n-1}\alpha^{2k+1}(x)+B\displaystyle\sum_{k=0}^{n-1}\beta^{2k+1}(x)\\ &=\dfrac{A\left(\alpha^{2n+1}(x)-\alpha(x)\right)}{\alpha^{2}(x)-1}+\dfrac{B\left(\beta^{2n+1}(x)-\beta(x)\right)}{\beta^{2}(x)-1}. \end{align*} Thus \begin{align*} \displaystyle\sum_{k=0}^{n-1}R_{2k+1}(x)&={\frac{A\alpha(x)+B\beta(x)-\alpha(x)\beta(x)(A\beta(x)+B\alpha(x))}{(\alpha(x)\beta(x))^2-\alpha^2(x)-\beta^2(x)+1}}\\ &\;\;\;-{\frac{A\alpha^{2n+1}(x)+B\beta^{2n+1}(x)+(\alpha(x)\beta(x))^2(A\alpha^{2n-1}(x)+B\beta^{2n-1}(x))}{(\alpha(x)\beta(x))^2-\alpha^2(x)-\beta^2(x)+1}}. \end{align*} Since \(\alpha(x)\beta(x)=-b\) and \(\alpha^2(x)+\beta^2(x)=a^2x^2+2b\) then using (8), (11) and (12), we obtain \begin{align*} \displaystyle\sum_{k=0}^{n-1}R_{2k+1}(x)&=\dfrac{R_{2n+1}(x)-b^2R_{2n-1}(x)+b{\left(R_1(x)-axR_0(x)\right)}-R_1(x)}{a^2x^2-b^2+2b-1}. \end{align*}
Proposition 4. [Sum of first \(n\) terms with even indices] The sum of first {n} terms of generalized Fibonacci sequences with even indices is given by \[\displaystyle\sum_{k=0}^{n-1} R_{2k}(x)=\dfrac{R_{2n}(x)-b^2R_{2n-2}(x)+(a^2x^2R_0(x)-axR_1(x)+bR_0(x))-R_0(x)}{a^2x^2-b^2+2b-1}.\]
Proof. Using Binet’s formula (8), we have \begin{align*} \displaystyle\sum_{k=0}^{n-1}R_{2k}(x)&=\displaystyle\sum_{k=0}^{n-1}{\left(A\alpha^{2k}(x)+B\beta^{2k}(x)\right)}=A\displaystyle\sum_{k=0}^{n-1}\alpha^{2k}(x)+B\displaystyle\sum_{k=0}^{n-1}\beta^{2k}(x)=\dfrac{A\left(\alpha^{2n}(x)-1\right)}{\alpha^2(x)-1}+{\dfrac{B\left(\beta^{2n}(x)-1\right)}{\beta^2(x)-1}}. \end{align*} Hence \begin{align*} \displaystyle\sum_{k=0}^{n-1}R_{2k}(x)&={\dfrac{A+B-(A\beta^2(x)+B\alpha^2(x))-(A\alpha^{2n}(x)+B\beta^{2n}(x))}{(\alpha(x)\beta(x))^2-\alpha^2(x)-\beta^2(x)+1}}+{\dfrac{(\alpha(x)\beta(x))^2(A\alpha^{2n-2}(x)+B\beta^{2n-2}(x))}{(\alpha(x)\beta(x))^2-\alpha^2(x)-\beta^2(x)+1}}. \end{align*} Since \(\alpha^2(x)+\beta^2(x)=a^2x^2+2b,\) and \(\alpha(x)\beta(x)=-b,\) then using (8), (10) and (13) , we obtain \[\displaystyle\sum_{k=0}^{n-1} R_{2k}(x)=\dfrac{R_{2n}(x)-b^2R_{2n-2}(x)+(a^2x^2R_0(x)-axR_1(x)+bR_0(x))-R_0(x)}{a^2x^2-b^2+2b-1}.\]
Proposition 5. For every positive integer \(n,\) we have \[\displaystyle\sum_{k=1}^{n}R_{3k}(x)=\dfrac{R_{3n+3}(x)+b^3R_{3n}(x)-R_{3}(x)-bR_0(x)}{a^3x^3+b^3+3abx-1}.\]
Proof. By Binet’s formula (8), we have \begin{align*} \displaystyle\sum_{k=1}^{n}R_{3k}(x)&=\displaystyle\sum_{k=1}^{n}{\left(A\alpha^{3k}(x)+B\beta^{3k}(x)\right)} =\dfrac{A\alpha^3(x)\left(\alpha^{3n}(x)-1\right)}{\alpha^3(x)-1}+{\dfrac{B\beta^3(x)\left(\beta^{3n}(x)-1\right)}{\beta^3(x)-1}}. \end{align*} Thus \begin{align*} \displaystyle\sum_{k=0}^{n}R_{3k}(x)&={\frac{(A\alpha^3(x)+B\beta^3(x))-(A\alpha^3\beta^3+B\beta^3\alpha^3)}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}\\ &\;\;\;+{\frac{(A\alpha^{3n+3}(x)\beta^3(x)+B\beta^{3n+3}(x)\alpha^3(x))-(A\alpha^{3n+3}(x)+B\beta^{3n+3}(x))}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}.\\ &={\frac{(A\alpha^3(x)+B\beta^3(x))-\alpha^3(x)\beta^3(x)(A+B)}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}\\ &\;\;\;+{\frac{\alpha^3(x)\beta^3(x)(A\alpha^{3n}(x)+B\beta^{3n}(x))-(A\alpha^{3n+3}(x)+B\beta^{3n+3}(x))}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}. \end{align*} Since \(\alpha^3(x)+\beta^3(x)=a^3x^3+3abx,\) and \(\alpha(x)\beta(x)=-b,\) then by Equations (8) and (10), we get \[\displaystyle\sum_{k=1}^{n} R_{3k}(x)=\dfrac{R_{3n+3}(x)+b^3R_{3n}(x)-R_3(x)-bR_0(x)}{a^3x^3+b^3+3abx-1}.\] Hence the proof.
Proposition 6. For every positive integer \(n,\) we have \[\displaystyle\sum_{k=1}^{n} R_{3k-1}(x)=\dfrac{R_{3n+2}(x)+b^3R_{3n-1}(x)+b^2(axR_0(x)-R_1(x))-R_{2}(x)}{a^3x^3+b^3+3abx-1}.\]
Proof. By Binet’s formula (8), we have \begin{align*} \displaystyle\sum_{k=1}^{n}R_{3k-1}(x)&=\displaystyle\sum_{k=1}^{n}{\left(A\alpha^{3k-1}(x)+B\beta^{3k-1}(x)\right)}=\dfrac{A\alpha^2(x)\left(\alpha^{3n}(x)-1\right)}{\alpha^3(x)-1}+{\dfrac{B\beta^2(x)\left(\beta^{3n}(x)-1\right)}{\beta^3(x)-1}}. \end{align*} This sum gives \begin{align*} \displaystyle\sum_{k=1}^{n}R_{3k-1}(x)&={\dfrac{A\alpha^2(x)+B\beta^2(x)-(A\alpha^2(x)\beta^3(x)+B\alpha^3(x)\beta^2(x))}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}\\ &\;\;\;+{\dfrac{A\alpha^{3n+2}(x)\beta^3(x)+B\beta^{3n+2}(x)\alpha^3(x)- A\alpha^{3n+2}(x)+B\beta^{3n+2}(x)}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}\\ &={\dfrac{A\alpha^2(x)+B\beta^2(x)-(\alpha(x)\beta(x))^2(A\beta(x)+B\alpha(x))}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}\\ &\;\;\;+{\dfrac{(\alpha(x)\beta(x))^3(A\alpha^{3n-1}(x)+B\beta^{3n-1}(x))- A\alpha^{3n+2}(x)+B\beta^{3n+2}(x)}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}. \end{align*} Since \(\alpha^3(x)+\beta^3(x)=a^3x^3+3abx,\) and \(\alpha(x)\beta(x)=-b,\) then making use of (8) and (12) , we obtain \[\displaystyle\sum_{k=1}^{n} R_{3k-1}(x)=\dfrac{R_{3n+2}(x)+b^3R_{3n-1}(x)+b^2(axR_0(x)-R_1(x))-R_{2}(x)}{a^3x^3+b^3+3abx-1}.\]
Proposition 7. For every positive integer \(n,\) we have \[\displaystyle\sum_{k=1}^{n} R_{3k-2}(x)=\dfrac{R_{3n+1}(x)+b^3R_{3n-2}(x)-b{\left(a^2x^2R_0(x)-axR_1(x)+bR_0(x)\right)-R_1(x)}}{a^3x^3+b^3+3abx-1}.\]
Proof. By Binet’s formula (8), we get \begin{align*} \displaystyle\sum_{k=1}^{n}R_{3k-2}(x)&=\displaystyle\sum_{k=1}^{n}{\left(A\alpha^{3k-2}(x)+B\beta^{3k-2}(x)\right)}\\ &=A\displaystyle\sum_{k=1}^{n}\alpha^{3k-2}(x)+B\displaystyle\sum_{k=1}^{n}\beta^{3k-2}(x)\\ &=\dfrac{A\alpha(x)\left(\alpha^{3n}(x)-1\right)}{\alpha^3(x)-1}+{\dfrac{B\beta(x)\left(\beta^{3n}(x)-1\right)}{\beta^3(x)-1}}. \end{align*} This Sum gives \begin{align*} \displaystyle\sum_{k=1}^{n}R_{3k-2}(x)&={\dfrac{A\alpha(x)+B\beta(x)-A\alpha\beta^3(x)+B\beta\alpha^3(x)}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}\\ &\;\;\;+{\dfrac{A\alpha^{3n+1}(x)\beta^3(x)+B\beta^{3n+1}(x)\alpha^3(x)- A\alpha^{3n+1}(x)+B\beta^{3n+1}(x)}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}\\ &={\dfrac{A\alpha(x)+B\beta(x)-\alpha(x)\beta(x)(A\alpha\beta^2(x)+B\alpha^2(x))}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}\\ &\;\;\;+{\dfrac{(\alpha(x)\beta(x))^3(A\alpha^{3n-2}(x)+B\beta^{3n-2}(x))- A\alpha^{3n+1}(x)+B\beta^{3n+1}(x)}{(\alpha(x)\beta(x))^3-\alpha^3(x)-\beta^3(x)+1}}. \end{align*} Since \(\alpha^3(x)+\beta^3(x)=a^3x^3+3abx,\) and \(\alpha(x)\beta(x)=-b,\) then making use of (8) and (13), we obtain \[\displaystyle\sum_{k=1}^{n} R_{3k-2}(x)=\dfrac{R_{3n+1}(x)+b^3R_{3n-2}(x)-b{\left(a^2x^2R_0(x)-axR_1(x)+bR_0(x)\right)-R_1(x)}}{a^3x^3+b^3+3abx-1}.\]
Theorem 3. [Generalized identity]Let \(R_n(x)\) be the \(n^{th}\) generalized Fibonacci polynomials. Then
Proof. Using Binet’s formula (8) to the left hand side, we have \begin{align*} \text{LHS}&={\left(A\alpha^m(x)+B\beta^m(x)\right)}{\left(A\alpha^n(x)+B\beta^n(x)\right)}-{\left(A\alpha^{m-k}(x)+B\beta^{m-k}(x) \right)}{\left(A\alpha^{n+k}(x)+B\beta^{n+k}(x)\right)}\\ &=AB{\left(\alpha^k(x)-\beta^k(x)\right)} {\left[\dfrac{\alpha^m(x)\beta^n(x)}{\alpha^k(x)}-\dfrac{\alpha^n(x)\beta^m(x)}{\beta^k(x)}\right]}\\ &=AB\dfrac{(\alpha^k(x)-\beta^k(x))}{(\alpha(x)\beta(x))^k}{\left(\alpha^m(x)\beta^{n+k}(x)-\alpha^{n+k}(x)\beta^m(x)\right)}\\ &=-AB(-b)^{m-k}{\left(\alpha^k(x)-\beta^k(x)\right)}{\left(\alpha^{n-m+k}(x)-\beta^{n-m+k}(x)\right)}. \end{align*} Since \(-AB=\dfrac{R_1^2(x)-R_0(x)R_2(x)}{(\alpha(x)-\beta(x))^2}\) by (14), then \begin{align*} &R_m(x)R_n(x)-R_{m-k}(x)R_{n+k}(x)=\dfrac{R_1^2(x)-R_0(x)R_2(x)}{(\alpha(x)-\beta(x))^2}(-b)^{m-k}{\left[(\alpha^k(x)-\beta^k(x))(\alpha^{n-m+k}(x)-\beta^{n-m+k}(x))\right]}\\ &={\left(R_1^2(x)-R_0(x)R_2(x)\right)}(-b)^{m-k}{\left[{\left(\dfrac{\alpha^k(x)-\beta^k(x)}{\alpha(x)-\beta(x)}\right)}{\left(\dfrac{\alpha^{n-m+k}(x)-\beta^{n-m+k}(x)}{\alpha(x)-\beta(x)}\right)}\right]}. \end{align*} From \[\dfrac{\alpha^k(x)-\beta^k(x)}{\alpha(x)-\beta(x)}=\dfrac{R_1(x)R_k(x)-R_0(x)R_{k+1}(x)}{R_1^2(x)-R_0(x)R_2(x)}\] and \[\dfrac{\alpha^{n-m+k}(x)-\beta^{n-m+k}(x)}{\alpha(x)-\beta(x)}=\dfrac{R_1(x)R_{n-m+k}(x)-R_0(x)R_{n-m+k+1}(x)}{R_1^2(x)-R_0(x)R_2(x)},\] we obtain our desired result.
Corollary 1. [Catalan’s identity] If \(m=n\) in the generalized identity (17), we obtain \[R_n^2(x)-R_{n-k}(x)R_{n+k}(x)=\dfrac{(-b)^{m-k}}{R_1^2(x)-R_0(x)R_2(x)}{\left[R_1(x)R_{k}(x)-R_0(x)R_{k+1}(x)\right]^2},\] where \(n>k\geq1.\)
Corollary 2. [Cassini’s identity] If \(m=n\) and \(k=1\) in the generalized identity (17), we obtain \[R_1^2(x)-R_{n-1}(x)R_{n+1}(x)=(-b)^{n-1}{\left[R_1^2(x)-R_0(x)R_2(x)\right]},\] for \(n\geq1.\)
Corollary 3. [d’Ocagne’s identity] If \(n=m,\) \(m=n+1\) and \(k=1\) in the generalized identity (17), we obtain \[R_m(x)R_{n+1}(x)-R_{m+1}(x)R_n(x)=(-b)^n{\left[R_1(x)R_{m-n}(x)-R_0(x)R_{m-n+1}(x)\right]},\] where \(m>n \geq1.\)