Inverse Fibonacci

Let's find the generating function for the fibbonacci series :

$F(x) = \sum_{n=1}^\infty F_n x^n.$

Using the fact that $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3$ , we get that:

$\begin{array} { l l } F(x) & = F_1x^1 + F_2x^2 + \sum_{n=3}^{\infty} ( F_{n-1} + F_{n-2} ) x^n \\ & = x + x^2 + \sum_{n=3}^{\infty} F_{n-1} x^n + \sum_{n=3}^{\infty} F_{n-2} x^n \\ & = x + x^2 + \left( -x^2 + \sum_{t=1}^\infty F_t x^{t+1} \right) + \left( \sum_{s=1}^\infty F_s x^{s+2 } \right) \\ & = x + x F(x) + x^2 F(x) \end{array}$

Thus, we get that $F(x) = \frac{x}{ 1 - x - x^2 }$ .

Now, substituting in $x = \frac{1}{10}$ , we obtain that

$a = \frac{1}{10} \times \sum_{i=1}^{\infty} \frac{ F_i} { 10^{i} } = \frac{1}{10} F ( \frac{1}{10} ) = \frac{1}{10} \times \frac{ \frac{1}{10} } { 1 - \frac{1}{10} - \left(\frac{1}{10} \right)^2 } = \frac{1}{89}.$

From this, the solutions to $a^2 x^2 - 2ax + 1 - a^2$ are $\{ \alpha, \beta \} = \{ 90, 88 \}$ , and so $| \alpha^2 - \beta^2| = 90^2 - 88^2 = 356$ .

$\frac {1}{100}+\frac {1}{1000}+\frac {2}{10000}+\frac {3}{100000}+... = \frac{1}{89}$

which leads to : $\alpha = 90 , \beta = 88$ and a final answer of $90^2-88^2=\boxed{\boxed {356}}$