Fibonacci equality, proving it someway

Question

$ F_{2n} = F_n(F_n+2F_{n-1}) $

$ F_n $ is a nth Fibonacci number. I tried by induction but i didn't get anywhere

Answer 1

A combinatorial interpretation:

$F_n$ is the number of ways to tile a row of $(n-1)$ squares with $1\times 1$ blocks and $1\times 2$ blocks.

The left hand side is the number of ways to tile a $1\times (2n-1)$ block with $1\times 1$ and $1\times 2$ blocks. Consider the middle square (the $(n-1)$th square.)

Case 1: It is used in a $1\times 1$ block. Then, there are $F_{n}$ ways to tile each of the $1 \times (n-1)$ blocks on each side of the middle, so $F_n^2$ total.

Case 2: It is used in a $1\times 2$ block. This block contains the $(n-1)$th square and either the $n$th or the $(n-2)$th square. In either case, there are $F_{n-1}$ ways to tile the shorter side and $F_n$ ways to tile the longer side.

We thus have $F_{2n} = F_n^2 + 2F_nF_{n-1},$ as desired.

Answer 2

This is very related to the answer at Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$

That question has the identity:

$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$

which can be modeled to your identity by letting $n=m$

$F_{2n} = F_{n-1}F_n + F_n F_{n+1}$

$F_{2n} = F_n(F_{n-1} + F_{n+1})$

Setting the expression inside the parenthesis to:

$F_{n-1} + F_{n+1} = F_{n-1}+ F_{n-1} + F_{n} = F_n + 2F_{n-1}$

We get

$F_{2n} = F_n(F_n+2F_{n-1})$

Which is your identity. So work backwards to that identity and use the proof at the linked question to prove your relation.

Answer 3

Here’s a purely computational proof. Let $$A=\pmatrix{F_2&F_1\\F_1&F_0}=\pmatrix{1&1\\1&0}\;;$$ a straightforward induction shows that $$A^n=\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}$$ for all $n\ge 1$. Then

$$\begin{align*} \pmatrix{F_{m+n+1}&F_{m+n}\\F_{m+n}&F_{m+n-1}}&=A^{m+n}\\ &=A^mA^n\\\\ &=\pmatrix{F_{m+1}&F_m\\F_m&F_{m-1}}\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}\\\\ &=\pmatrix{F_{m+1}F_{n+1}+F_mF_n&F_{m+1}F_n+F_mF_{n-1}\\F_mF_{n+1}+F_{m-1}F_n&F_mF_n+F_{m-1}F_{n-1}}\;, \end{align*}$$

so $F_{m+n}=F_{m+1}F_n+F_mF_{n-1}$. Take $m=n$, and this becomes

$$F_{2n}=F_{n+1}F_n+F_nF_{n-1}=F_n\left(F_{n+1}+F_{n-1}\right)=F_n\left(F_n+2F_{n-1}\right)\;.$$

Answer 4

An answer just by induction on $n$ of the equality $F_{2n}=F_n(F_{n-1}+F_{n+1})$ is as follows:

For $n=2$ we have $3=F_4=(1+3)\cdot 1=(F_1+F_3)F_2$.

To go from $n$ to $n+1$:

$$\begin{align} F_{2n+2}&=3F_{2n}-F_{2n-2}\\ &=3(F_{n-1}+F_{n+1})F_n-(F_{n-2}+F_n)F_{n-1}\\ &=F_{n-1}(3F_n-F_n-F_{n-2})+3F_{n+1}F_n\\ &=F_{n-1}(F_n+F_{n-1})+3F_{n+1}F_n\\ &=F_{n-1}F_{n+1}+3F_{n+1}F_n\\ &=F_{n+1}(3F_n+F_{n-1})\\ &=F_{n+1}(2F_n+F_{n+1})\\ &=F_{n+1}(F_n+F_{n+2}) \end{align}$$

Answer 5

This can be done also using the fact that

$$ F_n = \frac{1}{\sqrt{5}}\left(\sigma^n-\bar{\sigma}^n\right), $$

from which it is easy to get

$$ F_{2n} = F_n L_n, $$

(where $L_n$ is the $n$-th Lucas number) and we have only to prove

$$L_n= F_n+2F_{n-1},$$

that is true since $\{L_n\}_{n\in\mathbb{N}}$ and $\{F_n+2F_{n-1}\}_{n\in\mathbb{N}}$ are sequences with the same characteristic polynomial ($x^2-x-1$) and the same starting values $L_1=F_1+2F_0=1$, $L_2=F_2+2F_1=3$.

Answer 6

Another direct proof, using the fact that

$$ F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt5}\tag1 $$

where

$$ \phi=\frac{1+\sqrt5}{2},\qquad1-\phi=\frac{1-\sqrt5}{2}=-\frac{1}{\phi}. $$

From $(1)$ we have

$$ s_n\equiv\frac{F_{2n}}{F_n}=\frac{\phi^{2n}-(1-\phi)^{2n}}{\phi^n-(1-\phi)^n}=\phi^n+(1-\phi)^n $$

then

\begin{align} s_n-F_n&=\phi^n+(1-\phi)^n-\frac{\phi^n-(1-\phi)^n}{\sqrt5}=\\ &=\frac{1}{\sqrt5}\left[\sqrt5\phi^n+\sqrt5(1-\phi)^n-\phi^n+(1-\phi)^n\right]=\\ &=\frac{1}{\sqrt5}\left[-(1-\sqrt5)\phi^n+(1+\sqrt5)(1-\phi)^n\right]=\\ &=\frac{1}{\sqrt5}\left[-2(1-\phi)\phi^n+2\phi(1-\phi)^n\right]=\\ &=-\frac{2\phi(1-\phi)}{\sqrt5}\left[\phi^{n-1}-(1-\phi)^{n-1}\right]=2F_{n-1} \end{align}