Prove Lucas numbers and Fibonacci numbers relation $F_n = \frac{L_{n - 1} + L_{n + 1}}{5}$ $F_n$, is the $n$th term of the Fibonacci sequence.
$L_n$ is the $n$th Lucas number.
I want to prove that $F_n = \dfrac{L_{n-1}+L_{n+1}}5 $.
Things I know:


*

*$L_n$+$L_{n+1}=L_{n+2}$ 

*$F_{n-1} + F_{n+1}=L_n$.


The base case is easy to prove. I will omit proving the base case it is simple.
My attempt:
$F_{n+1} = \dfrac{L_{n}+L_{n+2}}5 $.
Attempt one :using and manipulating $#2$ make all $L_n$ in terms of $F_n$. 
Attempt two:
Using $#2$ two make $F_{n+1}$ in terms of $L_n$. 
 A: The Fibonacci numbers are defined recursively by 
$$
F_n = 
\begin{cases}
1 & \text{if $n = 1$}\\
1 & \text{if $n = 2$}\\
F_{n - 1} + F_{n - 2} & \text{if $n > 2$}
\end{cases}
$$
and the Lucas numbers are defined recursively by 
$$
L_n =
\begin{cases}
2 & \text{if $n = 0$}\\
1 & \text{if $n = 1$}\\
L_{n - 1} + L_{n - 2} & \text{if $n > 1$}
\end{cases}
$$
Proof.  Let $P(n)$ the statement 
$$F_n = \frac{L_{n + 1} + L_{n - 1}}{5}$$
We will prove $P(n)$ holds for all positive integers $n$ using strong induction.  
Let $n = 1$.  Observe that $L_2 = L_1 + L_0 = 1 + 2 = 3$. Thus, 
$$\frac{L_{1 + 1} + L_{1 - 1}}{5} = \frac{L_2 + L_0}{5} = \frac{3 + 2}{5} = \frac{5}{5} = 1 = F_1$$
so $P(1)$ holds.
Let $n = 2$.  Observe that $L_3 = L_2 + L_1 = 3 + 1 = 4$.  Thus,
$$\frac{L_{2 + 1} + L_{2 - 1}}{5} = \frac{L_3 + L_1}{5} = \frac{4 + 1}{5} = \frac{5}{5} = 1 = F_2$$
so $P(2)$ also holds.
Assume $P(n)$ holds for each positive integer $n \leq m$, where $m \geq 2$.  Then 
$$F_m = \frac{L_{m + 1} + L_{m - 1}}{5}$$
and 
$$F_{m - 1} = \frac{L_{(m - 1) + 1} + L_{(m - 1) - 1}}{5} = \frac{L_m + L_{m - 2}}{5}$$
Let $n = m + 1$.  Then
\begin{align*}
\frac{L_{(m + 1) + 1} + L_{(m + 1) - 1}}{5} & = \frac{L_{m + 2} + L_{m}}{5}\\
& = \frac{L_{m + 1} + L_m + L_{m - 1} + L_{m - 2}}{5}\\
& = \frac{L_{m + 1} + L_{m - 1} + L_m + L_{m - 2}}{5}\\
& = \frac{L_{m + 1} + L_{m - 1}}{5} + \frac{L_m + L_{m - 2}}{5}\\
& = F_m + F_{m - 1}\\
& = F_{m + 1}
\end{align*}
Thus, $P(n)$ holds for each positive integer $n$.$\blacksquare$
