another look at Fibonacci identity (can't use induction)

I recently submitted this question (here)with help from this board by using induction and my professor told me to try to solve it a different way and hinted at using the binomial theorem.

I know that the Binomial Theorem is:

(x+y)^2=(n C 0)x^n + (n C 1)x^n-1 * y + (n C 2)x^n-2 * y^2....(n C n)y^n

but I'm not sure where to begin. The question again is:

Prove the identity where Fn is the n-th Fibonacci number.

[n 0)*F0 + (n 1)*F1 + (n 2)*F2 + (n 3)*F3 +. . .+ (n n)*Fn = F2n

(n 0) = n choose 0...etc
and F0 = F sub 0...etc

Found this identity:

\(\displaystyle \L F_n = \sum_{k=1}^n { n-k \choose k-1 }\)