TITLE: Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$
QUESTION [8 upvotes]: Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$; 
I was stuck with this question for a while... Help me please!!! Thanks!!!

REPLY [6 votes]: I accidentally stumbled across this old question while studying some facts about generating functions and could not resist posting this answer.
Take an ordinary generating function of the LHS:
$$G(x)=\sum_{n\ge0}\sum_{0\le k\le n}\left(\begin{array}{c}
n\\
k
\end{array}\right)F_kx^n$$
Change the order of summation to obtain
$$G(x)=\sum_{k\ge0}\sum_{n\ge k}\left(\begin{array}{c}
n\\
k
\end{array}\right)F_kx^n=\sum_{k\ge0}F_k\sum_{n\ge0}\left(\begin{array}{c}
n+k\\
k
\end{array}\right)x^{n+k}\\=\sum_{k\ge0}F_kx^k\sum_{n\ge0}\left(\begin{array}{c}
n+k\\
n
\end{array}\right)x^n=\sum_{k\ge0}F_kx^k\frac{1}{\left(1-x\right)^{1+k}}\\=\frac{1}{1-x}\sum_{k\ge0}F_k\left(\frac{x}{1-x}\right)^k$$
Now the expression under the sum is just the ordinary generating $F(z)$function for $F_n$
$$F(z)=\frac{z}{1-z-z^2}$$
where $z=\frac{x}{1-x}$. Substituting we obtain
$$G(x)=\frac{x}{1-3x-x^2}$$
The last expression is the generating function for $F_{2n}$ as can be ascertained by calculating $\frac{1}{2}\left[F(x)+F(-x)\right]$