$F_n$, 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 $L_n$.
Best Answer
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 $n$.$\blacksquare$