Inspired by Otto...Funny back to the future

$\displaystyle \sum_{n=1}^{2013}{f_n f_{n+1}} = \sum_{n=1}^{2013}{(f_{n+1}-f_{n-1})f_{n+1}}$

$\displaystyle \sum_{n=1}^{2013}{f_{n+1}^2 - f_{n-1}f_{n+1}}$

$\text{Lemma :} f_n f_{n+2} = f_{n+1}^2 + (-1)^n$

Using this,

$\displaystyle \sum_{n=1}^{2013}{f_{n+1}^2 - f_n^2 - (-1)^n}$

Now, this easily telescopes to

$\displaystyle f_{2014}^2 - 1 + 1 = \boxed{f_{2014}^2}$

$1 \cdot 1$ = $1^2$ = $f_1 \cdot f_2$

$1 \cdot 1$ + $1 \cdot 2$ + $2 \cdot 3$ = $3^2$ = $f_1 \cdot f_2$ + $f_2 \cdot f_3$ + $f_3 \cdot f_4$

$1 \cdot 1$ + $1 \cdot 2$ + $2 \cdot 3$ + $3 \cdot 5$ + $5 \cdot 8$ = $8^2$ = $f_1 \cdot f_2$ + $f_2 \cdot f_3$ + $f_3 \cdot f_4$ + $f_4 \cdot f_5$ + $f_5 \cdot f_6$

Using induction and seeing the picture above $(1 \cdot 1)$ + $(1 \cdot 2)$ + $(2 \cdot 3)$ + $(3 \cdot 5)$ + ... + $(f_{2012} \cdot f_{2013})$ + $(f_ {2013} \cdot f_{2014})$ = $f^2_{2014}$ $\Rightarrow$ n=2014

The expression is basically:

$f_{1})(f_{2})+(f_{2})(f_{3})+(f_{3})(f_{4})+......(f_{2011})(f_{2012})+(f_{2012})(f_{2013})+(f_{2013})(f_{2014})=f^2_{n}$

Which factors to;

$f_{2}(f_{1}+f_{3})+f_{4}(f_{3}+f_{5})+f_{6}(f_{5}+f_{7})+.....f_{2012}(f_{2011}+f_{2013})+(f_{2013})(f_{2014})=f^2_{n}$

Based on the definition of Fibonacci sequence, $f_{n}=f_{n-2}+f_{n-1}$ ,

We have the related expression

$(f_{3}-f_{1})(f_{1}+f_{3})+(f_{5}-f_{3})(f_{3}+f_{5})+.......(f_{2013}-f_{2011})(f_{2011}+f_{2013})+(f_{2013})(f_{2014})=f^2_{n}$

Which equals to:

$(f^2_{3}-f^2_{1})+(f^2_{5}-f^2_{3})+(f^2_{7}-f^2_{5})+......(f^2_{2013}-f^2_{2011})+(f_{2013})(f_{2014})=f^2_{n}$

All the middle terms cancel out except for the $f^2_{2013}$ :

$-f^2_{1}+f^2_{2013}+(f_{2013}*f_{2014})=f^2_{n}$

Which equals:

$-1+f_{2013}(f_{2013}+f_{2014})=f^2_{n}$

Which is the same as:

$-1+f_{2013}(f_{2015})=f^2_{n}$ (eq.1)

Using the identityof Fibonacci sequence $f_{n+1}(f_{n-1})-f^2_{n}=(-1)^n$ and let $n=2014$ :

$f_{2013}(f_{2015})-f^2_{2014}=(-1)^{2014}=1$

Thus $f_{2013}(f_{2015})=1+f^2_{2014}$

Substitute into the original equation:

$-1+1+f^2_{2014}=f^2_{n}$