Without resorting to induction show that $L_n^2=L_{n+1}L_{n-1}+5(-1)^n$,Where $L_n$ is $n^{th}$ Lucas number.

Question

Without resorting to induction show that $L_n^2=L_{n+1}L_{n-1}+5(-1)^n$,Where $L_n$ is $n^{th}$ Lucas number.

By definition of Lucas number $L_n=L_{n-1}+L_{n-2}\implies L_{n-1}=L_n-L_{n-2}$ and $L_{n+1}=L_{n}+L_{n-2}.$

Now,$L_{n+1}L_{n-1}=L_n^2+L_{n}L_{n-1}-L_{n}L_{n-2}-L_{n-2}L_{n-1}$

From,here i'm not getting how to proceed further...

Answer 1

The identity $$ L_n^2=L_{n+1}L_{n-1}+5(-1)^n $$ for the Lucas numbers seems to be closed related to Cassini's identity for the Fibonacci numbers: $$ F_n^2 - F_{n+1}F_{n-1} = (-1)^{n-1} $$ which follows from taking determinants in $$ \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n $$

Indeed, the identity in question has a closely related proof: it follows from taking determinants in $$ \begin{pmatrix}L_{n+1}&L_n\\L_n&L_{n-1}\end{pmatrix} = \begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1} \begin{pmatrix}L_{2}&L_1\\L_1&L_{0}\end{pmatrix} = \begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1} \begin{pmatrix}3&1\\1&2\end{pmatrix} $$

Answer 2

Using the Binet formula, $$L_n=a^n+b^n$$ where $a,b$ are the roots of $$t^2-t-1=0$$

$\implies ab=-1, a+b=1\implies a^2+b^2=(a+b)^2-2ab=?$

$$\implies L_n^2-L_{n-1}L_{n+1}$$

$$=(a^n+b^n)^2-(a^{n+1}+b^{n+1})(a^{n-1}+b^{n-1})$$

$$=2(ab)^n-(ab)^{n-1}(a^2+b^2)=?$$