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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 :-)

Fill
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    There is no differeence between "show that" and "prove that" and induction is probably the right tool. However you have a small error. The sum has $2n$ terms. So when you increase $n$ by one you should have 2 leftover terms. not just one. – Jorik Nov 13 '15 at 17:33
  • which means I have to add $f_{2n}f_{2n+1}+f_{2n+1}f_{2n+2}$ to both sides? – Fill Nov 13 '15 at 17:37

1 Answers1

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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.

SchrodingersCat
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