More fun in 2015, Part 16

Apply Binet's formula http://www.artofproblemsolving.com/wiki/index.php/Binet%27s_Formula And note that the series now becomes a sum of two infinite geometric series, which can be easily evaluated.

(I am on my phone as of the time of writing so latex is a pain, but I can offer elaboration later if someone wants me to. The key step here though is application of Binet's, and the rest should be algebra.)

EDIT: here it is.

$\sum_{n=0}^\infty \frac{F_n}{2015^n} = \sum_{n=0}^\infty \frac{\frac{1}{\sqrt5}\big(\frac{1+\sqrt5}{2}\big)^n - \frac{1}{\sqrt5}\big(\frac{1 - \sqrt5}{2}\big)^n}{2015^n}$ (by Binet's) $= \sum_{n=0}^\infty \frac{1}{\sqrt5}\big(\frac{1+\sqrt5}{2\cdot 2015}\big)^n - \sum_{m=0}^\infty\frac{1}{\sqrt5} \big(\frac{1 - \sqrt5}{2\cdot 2015}\big)^m$ Which is a difference of two geometric series. Evaluating each of them, the above is equivalent to
$\frac{1}{\sqrt5}\frac{1}{1 - \frac{1+\sqrt5}{4030}} - \frac{1}{\sqrt5}\frac{1}{1 - \frac{1+\sqrt5}{4030}}$ $= \frac{1}{\sqrt5}\Big(\frac{4030}{4029 - \sqrt5} - \frac{4030}{4029 - \sqrt5}\Big)$ $= \frac{1}{\sqrt5}\Big(\frac{4030(4029 + \sqrt5) - 4030(4029 + \sqrt5)}{4029^2 - 5} \Big)$ $= \frac{1}{\sqrt5}\Big(\frac{2\cdot2\cdot2015\sqrt5}{4029^2 - 5} \Big)$ $=\frac{2\cdot2\cdot2015\sqrt5}{\sqrt5(4029^2 - 5)}$ $= \frac{2015}{4058209}$ so our answer is $\boxed{2015}$ and wow that was kinda messy

My solution is the same as @Jake Lai's solution to Fibonacci is fun :

$\sum^{\infty}_{n=0} {F_n}{x^n}=\frac{x}{1-x-x^2}$ for all |x|< $\frac{1}{\phi}\approx0.618$ , where $\phi$ is the golden ratio. Clearly $\frac 1 {2015}$ is less than $\frac 1 {\phi}$ .

Substituting x= $\frac 1 {2015}$ in the expression, we can see that $a = \boxed{2015}$ .

Jake Lai has also written other solutions to these kinds of problems which use basic algebra to arrive at the answer. I encourage you to check out his solutions to the "Fibonacci is fun" problem.

The generating function for the Fibonacci numbers is

g(x) = sum (n=0)^(infty)F nx^n

=   x/(1-x-x^2)

Substituting the values x = 1/2015 gives us , a = 2015 and b = 4058209

This problem could have been made harder by asking the viewer to determine the value of b instead of a