Proof involving fibonacci number: $ \binom{n}{1}F_1+\binom{n}{2}F_2+\binom{n}{3}F_3+\cdots+\binom{n}{n-1}F_{n-1}+F_n=F_{2n}, $
Fibonacci combinatorial identity: $F_{2n} = {n \choose 0} F_0 + {n\choose 1} F_1 + ... {n\choose n} F_n$