Madhava from Sangamagrama?

This is only my humble approach from "thundering" Otto Bretscher's solution.

We are considering the function $\arctan : (-\infty,\infty) \rightarrow (\frac{-\pi}{2},\frac{\pi}{2})$ as an inverse function of a bijective function strictly increasing( $\tan$ )

1.- Let $c_n = \arctan(\frac{1}{F_n})$ and firstly we are going to see this sequence $\displaystyle \sum_{n = 1}^\infty c_{2n + 1}$ converges. For it, I'm going to use Taylor series of $\arctan (x)$ ~ $x$ as $x \to 0$ and D'Alembert convergence criterium $\displaystyle \lim_{n \to \infty} \frac {c_{2n + 3}}{c_{2n + 1}} = \lim_{n \to \infty} \frac {F_{2n + 1}}{F_{2n + 3}} = \lim_{n \to \infty} \frac {F_{2n + 1}}{F_{2n + 2}} \cdot \frac{F_{2n + 2}}{F_{2n + 3}} = \frac{1}{\phi^2} < 1 \Rightarrow$ the sequence $\displaystyle \sum_{n = 1}^\infty c_{2n + 1}$ converges,where $\phi$ is the golden ratio. This means the series $\displaystyle \sum_{n = 1}^\infty c_{2n + 1} x^n$ has radius of convergence $\rho = \phi^2$ ,i.e, the series converges if $|x| < \rho = \phi^2$ and $x \in \mathbb{R}$ , even if $x \in \mathbb{C}$ and $|x| < \rho = \phi^2$

2.- The second part is only a translation from Otto's, using Cassini's identity ( $F_{n + i}F_{n + j} - F_n F_{n + i + j} = (-1)^n F_i F_j$ ) $F_{2n - 1}F_{2n + 2} - 1 = F_{2n - 1}F_{2n + 2} + (-1)^{2n - 1} F_1 F_ 2 = F_{2n}F_{2n + 1} \iff$ $\iff (F_{2n - 1} + F_{2n})F_{2n + 2} - 1 = F_{2n}(F_{2n + 1} + F_{2n + 2}) \iff$ $F_{2n + 1}F_{2n + 2} - 1 = F_{2n}(F_{2n + 1} + F_{2n + 2}) \iff$ $\iff \frac{1}{F_{2n}} = \frac{F_{2n + 1} + F_{2n + 2}}{F_{2n + 1}F_{2n + 2} - 1} \iff \tan(c_{2n}) = \tan ( c_{2n +1} + c_{2n + 2}) \iff c_{2n} = c_{2n + 1} + c_{2n + 2}.$ The rest of proof is exactly equal to Otto's $c_3 = c_2 - c_4 \\ c_5 = c_4 - c_6 \\ c_7 = c_6 - c_8...$

Another lovely problem, compañero!

Let $c_n=\arctan(\frac{1}{F_n})$ . The identity $c_{2n}=c_{2n+1}+c_{2n+2}$ follows from the addition theorem for tan and Cassini's identity. By induction we find $\frac{\pi}{4}=c_2=c_{2N+2}+\sum_{n=1}^{N}c_{2n+1}$ . The limit as $N$ tends to infinity is $\frac{\pi}{4}=\sum_{n=1}^{\infty}c_{2n+1}$ so that the answer is $10\times1+4=\boxed{14}$