How to show that $f_{0}f_{1}+f_{1}f_{2}+...+f_{2n-1}f_{2n} = f^{2}_{2n}$ The following problem:
The Fibonacci number is denoted by  $f_{n}$, show that the following holds when $n$ is a positive integer:
$f_{0}f_{1}+f_{1}f_{2}+...+f_{2n-1}f_{2n} = f^{2}_{2n}$
My idea was to do an induction proof, which holds for $n$ = 1
$f_{0}f_{1} + f_{1}f_{2} = f^{2}_{2}$
So we assume it holds for $n$ and show for $n+1$
$f_{0}f_{1}+f_{1}f_{2}+...+f_{2n-1}f_{2n} + f_{2n}f_{2n+1} = f^{2}_{2n} + f_{2n}f_{2n+1}$
I fail to get  $f^{2}_{2n+1}$ out of the term  $f^{2}_{2n} + f_{2n}f_{2n+1}$
Is there a difference between "show that" and "prove that". So might be induction the wrong tool here?
Many thanks for your hints in advance :-)

The following step has a mistake as Jorik has mentioned.

So we assume it holds for $n$ and show for $n+1$
$$f_{0}f_{1}+f_{1}f_{2}+...+f_{2n-1}f_{2n} + f_{2n}f_{2n+1} = f^{2}_{2n} + f_{2n}f_{2n+1}$$

Rather it should be as follows:
$$f_{0}f_{1}+f_{1}f_{2}+...+f_{2n-1}f_{2n} + f_{2n}f_{2n+1} + f_{2n+1}f_{2n+2}$$
$$= f^{2}_{2n} + f_{2n}f_{2n+1} + f_{2n+1}f_{2n+2}$$
$$= f^{2}_{2n} + f_{2n}f_{2n+1} + f_{2n+1}(f_{2n+1}+f_{2n})$$ using the property of Fibonacci numbers
$$= f^{2}_{2n} + 2f_{2n}f_{2n+1} + f^{2}_{2n+1}$$
$$= (f_{2n}+f_{2n+1})^2$$
$$= f^{2}_{2n+2}$$
$$= f^{2}_{2(n+1)}$$
Hope it helps.
