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$F_n$, 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:

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

N. F. Taussig
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hajeet
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  • how about using the explicit expression for both Fibonacci and Lucas (the latter I assume it exists), and you check the equality directly? – Chip Feb 22 '16 at 05:21
  • Prove two base cases. Have two hypothesis steps, $P(k-1)$ and $P(k)$. Then, add both of them to get P(k+1)$ – Saikat Feb 22 '16 at 13:10
  • @Chip An explicit formula for the Fibonacci numbers is $$F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$ for $n \geq 1$, where $$\varphi = \frac{1 + \sqrt{5}}{2}$$ and $$\psi = \frac{1 - \sqrt{5}}{2}$$ An explicit formula for the Lucas numbers is $$L_n = \varphi^n + \psi^n$$ for $n \geq 0$, where $\varphi$ and $\psi$ are defined as above. – N. F. Taussig Feb 24 '16 at 14:17

1 Answers1

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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 $n$.$\blacksquare$

N. F. Taussig
  • 79,074