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. 
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...

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
$$
\begin{pmatrix}L_{n+1}&L_n\\L_n&L_{n-1}\end{pmatrix}
=
\begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1}
\begin{pmatrix}L_{2}&L_1\\L_1&L_{0}\end{pmatrix}
=
\begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1}
\begin{pmatrix}3&1\\1&2\end{pmatrix}
$$
