I really feel like I am missing something.
Geometricly, this sum is clear:
it's $f_{2n+1}-1$
$$\require{cancel}$$
But when I use the identity $f_{2n}=f_{n+1}^2-f_{n-1}^2$ I get the following $$0+(f_2^2-f_0^2)+(f_3^2-f_1^2)+(f_4^2-f_2^2)+\cdots+(f_{n+1}^2-f_{n-1}^2)=$$ $$=0+(\cancel{f_2^2}-f_0^2)+(\cancel{f_3^2}-f_1^2)+(\cancel{f_4^2}-\cancel{f_2^2})+\cdots+(f_{n+1}^2-\cancel{f_{n-1}^2})=$$ $$=f_{n+1}^2-1$$ $$\therefore f_{n+1}^2-1=f_{2n+1}-1$$ $$f_{n+1}^2=f_{2n+1}$$ I can also use $f_{2n+1}=f_{n+1}^2+f_n^2$: $$f_{n+1}^2=f_{n+1}^2+f_n^2$$ $$\Rightarrow0=f_n^2$$ Which is clearly wrong. Where did I mess up?
You’ve canceled the terms incorrectly I think. The last 2 terms will be $$(f_n^2\ -\ f_{n-2}^2)\ +\ (f_{n+1}^2\ -\ f_{n-1}^2)$$ Out of which, $$ f_n^2 \ and\ f_{n+1}^2 $$ won’t cancel out. Thus eventually you’ll get $0=0$