Fibonacci Proof 
Prove that: $$F_1F_2+F_2F_3+F_3F_4+\cdots+F_{2n-1}F_{2n}=F_{2n}^2$$

I set it up so:
$$F^2(2k) + F(2k+1)F(2k+2) = F^2(2k+2)$$
I've tried:
$$F(2k)^2 + F(2k+1)*F(2k+2) = F(2k+2)*F(2k)+F(2k+2)*F(2k+1)$$
$$F(2k)^2 = F(2k+2)*F(2k)$$
$$F(2k+1)*F(2k) - F(2k)*F(2k-1) = F(2k)*F(2k)+F(2k)*F(2k+1)$$
$$-F(2k)*F(2k-1) = F(2k)^2$$
??? Is there something wrong with my initial induction hypothesis ? 

$$F_{r+2}^2-F_r^2=(F_{r+2}-F_r)(F_{r+2}+F_r)=F_{r+1}(F_{r+2}+F_r)$$
So, $$F_{r+2}^2-F_r^2=F_{r+2}F_{r+1}+F_{r+1}F_r$$ 
We can utilize this fact to prove the required  proposition without using induction.
Put $r=2n-2,2n-4,\cdots,4,2,0$ to get,
$$F_{2n}^2-F_{2n-2}^2=F_{2n}F_{2n-1}+F_{2n-1}F_{2n-2}$$ 
$$F_{2n-2}^2-F_{2n-4}^2=F_{2n-2}F_{2n-3}+F_{2n-3}F_{2n-4}$$ 
$$\cdots$$
$$F_{6}^2-F_{4}^2=F_{6}F_{5}+F_{5}F_{4}$$ 
$$F_{4}^2-F_{2}^2=F_{4}F_{3}+F_{3}F_{2}$$ 
$$F_{2}^2-F_{0}^2=F_{2}F_{1}+F_{1}F_{0}$$ 
Adding them we get, $$F_{2n}^2-F_0^2=\sum_{0\le m\le 2n-1}F_mF_{m+1}$$
But $$F_0=0\implies  F_{2n}^2-F_0^2=F_{2n}^2$$
 and  $$ \sum_{0\le m\le 2n-1}F_mF_{m+1}=\sum_{1\le m\le 2n-1}F_mF_{m+1}+F_0F_1=\sum_{1\le m\le 2n-1}F_mF_{m+1}$$
So, $$F_{2n}^2=\sum_{1\le m\le 2n-1}F_mF_{m+1}$$
Similarly putting $r=2n-1,2n-3,\cdots,3,1$ and adding them we can derive $$\sum_{1\le m\le 2n}F_mF_{m+1}=F_{2n+1}^2-F_1^2=F_{2n+1}^2-1\text{ as }F_1=1$$
