Is there a Fibonacci identity? Let $F(n)$ be the n-th Fibonacci number. Is there any identity to simplify the following summation.
$$\sum_{i=1}^{n-1}F(i)^2F(n-i)^2 $$
 A: The first few terms are $1,2,9,26,84$.  Putting this into OEIS gives A136429.  There you can find a generating function, an expression as a self-convolution, and the first 100 terms.  Also, they satisfy the recurrence
$$a_{n+6}=4a_{n+5}-10a_{n+3}+4a_{n+1}-a_n$$
In principle, you can find a closed form from this, but that would require finding all the roots to a sixth-degree polynomial.  Luckily, that polynomial isn't too difficult, and has roots $$-1, \frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}$$
Hence the general sequence satisfying the above recurrence is $$a_n=A(-1)^n+B(\frac{3+\sqrt{5}}{2})^n+Cn(\frac{3+\sqrt{5}}{2})^n+D(\frac{3-\sqrt{5}}{2})^n+En(\frac{3-\sqrt{5}}{2})^n$$
You can substitute the known five initial values, above, for $a_2$ through $a_6$, which will lead to a system of five linear equations in the five unknowns $A,B,C,D,E$.  Then you may solve this system to get a closed form for $a_n$.   See Robert Israel's worked out closed form in the comments.
A: $$F(i)=\frac{a^i-b^i}{a-b},\ a=\frac{1+\sqrt{5}}{2},\ b=\frac{1-\sqrt{5}}{2},\\ F^2(i)=\frac{a^{2i}+b^{2i}-2(-1)^i}{5}\implies S(x)=\sum_{i\ge 1}F^{2i}x^i=\frac{\sum_{i\ge 1}a^{2i}x^i+\sum_{i\ge 1}b^{2i}x^i-2\sum_{i\ge 1}(-x)^i}{5}=\frac{1}{5}\left(\frac{1}{1-a^2x}+\frac{1}{1-b^2x}-\frac{2}{1+x}\right)=\frac{1}{5}\left(\frac{x^2+5x}{1-2x-2x^2+x^3}\right)$$ Then the given expression will be the coefficient of $x^{n}$ in the expansion of $S^2(x)$
