<s>[INST] 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 ? 
 [/INST] We can avoid induction using Binet's Fibonacci Number Formula, $$F_n=\frac{a^n-b^n}{a-b}$$ where $a,b$ are the roots of $x^2-x-1=0\iff a+b=1,ab=-1$ also, $a^2-a-1=0$
$$F_rF_{r+1}=\frac{(a^r-b^r)}{(a-b)}\frac{(a^{r+1}-b^{r+1})}{(a-b)}=\frac{a^{2r+1}+b^{2r+1}-(ab)^r(a+b)}{(a-b)^2}$$
$$=\frac{a^{2r+1}+b^{2r+1}-(-1)^r}{(a-b)^2} \text{  as } ab=-1$$
So, $$(a-b)^2\sum_{m\le r\le m+n}F_rF_{r+1}$$
$$=\sum_{m\le r\le m+n}a^{2r+1}+\sum_{m\le r\le m+n}b^{2r+1}-\sum_{m\le r\le m+n}(-1)^r$$
$$=\frac{a^{2m+1}\{(a^2)^{n+1}-1\}}{a^2-1}+\frac{b^{2m+1}\{(b^2)^{n+1}-1\}}{b^2-1}-(-1)^m\sum_{0\le r\le n}(-1)^r$$
$$=a^{2(m+n+1)}+b^{2(m+n+1)}-(a^{2m}+b^{2m})-(-1)^m\sum_{0\le r\le n}(-1)^r$$
as $a^2-1=a$ as $a^2-a-1=0$
$$=(a^{m+n+1}-b^{m+n+1})^2+2(ab)^{m+n+1}-\{(a^m-b^m)^2+2(ab)^m\}-(-1)^m\sum_{0\le r\le n}(-1)^r$$
$$=(a^{m+n+1}-b^{m+n+1})^2-(a^m-b^m)^2-2(-1)^m\{(-1)^{n+1}-1\}-(-1)^m\sum_{0\le r\le n}(-1)^r$$
As the sum of even number of alternative positive negative real terms with same absolute value is $0,$ if $n$ is odd $=2k-1$(say),
$$(a-b)^2\sum_{m\le r\le m+2k-1}F_rF_{r+1}=(a^{m+2k}-b^{m+2k})^2-(a^m-b^m)^2$$
$$\implies \sum_{m\le r\le m+2k-1}F_rF_{r+1}=F_{m+2k}^2-F_m^2$$
$$m=0\implies \sum_{0\le r\le 2k-1}F_rF_{r+1}=F_{2k}^2-F_0^2=F_{2k}^2 \text{ as } F_0=0$$
$$\implies \sum_{1\le r\le 2k-1}F_rF_{r+1}=F_{2k}^2-F_0F_1=F_{2k}^2 \text{ as } F_0=0$$
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