TITLE: Counting argument proof or inductive proof of $F_1 {n \choose1}+...+F_n {n \choose n} = F_{2n}$ where $F_i$ are Fibonacci
QUESTION [8 upvotes]: Prove that $$F_1 {n\choose1} +...+F_n {n\choose n} = F_{2n}$$
where $F_i$ is the $i$th Fibonacci number.
This question is mentioned on Putnam and Beyond book number 861. I looked at the proof of the question, and the book's proof uses Binet's formula (i.e. the explicit formula for $n$th Fibonacci number), binomial formula, and the fact that $\frac{3+\sqrt{5}}{2} = (\frac{1+\sqrt{5}}{2})^2$. I am wondering whether there is a counting argument for this proof (or an inductive proof), and if so, it would be great if anyone posts them here.

REPLY [4 votes]: Although this not my answer, I would like to rewrite it here. In fact, every time I read the @Robert Israel answer I understand that I  know a bit about Fibonacci numbers and not more! 
Consider $A=\pmatrix{0 & 1\cr 1 & 1\cr}$ then the $n$th power of $A$, denoted by $A^n$, is obtained as follows 
$$
A^n = \pmatrix{F_{n-1} & F_{n}\cr F_n & F_{n+1}} .
$$
where $F_n$ is the $n$th term of Fibonacci numbers. 
The matrix $A$ is satisfaied on the equation $x^2-x-1=0$, since we have 
$$
 \pmatrix{1 & 1\cr 1 & 2\cr}=\pmatrix{0 & 1\cr 1 & 1\cr}+\pmatrix{1 & 0\cr 0 & 1\cr} \Leftrightarrow  A^2=A+I 
$$
Therefore,  we get 
$$
A^{2n}=(A^2)^n=(A+I)^n=\sum_{j=0}^n C_j^n\, A^j \tag{1}
$$
Now consider the entry of the first row and second column of the matrix equation $(1)$ which impliying that
$$
F_{2n}=\sum_{j=0}^n C_j^n\, F_j\, .
$$