Fibonacci equality, proving it someway $ 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
 A: 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.
A: 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. 
A: 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
$$F_{2n}=F_{n+1}F_n+F_nF_{n-1}=F_n\left(F_{n+1}+F_{n-1}\right)=F_n\left(F_n+2F_{n-1}\right)\;.$$
A: 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 $n+1$:
$$\begin{align}
F_{2n+2}&=3F_{2n}-F_{2n-2}\\
&=3(F_{n-1}+F_{n+1})F_n-(F_{n-2}+F_n)F_{n-1}\\
&=F_{n-1}(3F_n-F_n-F_{n-2})+3F_{n+1}F_n\\
&=F_{n-1}(F_n+F_{n-1})+3F_{n+1}F_n\\
&=F_{n-1}F_{n+1}+3F_{n+1}F_n\\
&=F_{n+1}(3F_n+F_{n-1})\\
&=F_{n+1}(2F_n+F_{n+1})\\
&=F_{n+1}(F_n+F_{n+2})
\end{align}$$
A: 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 $L_1=F_1+2F_0=1$, $L_2=F_2+2F_1=3$.
A: 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}
