I Have More Bananas!

Here is a sketch.

$f'(x) = \sum_{n=0}^{\infty} a_{n}x^{n}=\frac{2x+1}{1-x-x^2}$

$f(x) = -\ln(1-x-x^2)$

Plugging $x=\frac{1}{2}$ in we get $f(\frac{1}{2})=\ln(4)$ $\implies k=\boxed{4}$

Since the Lucas sequence is a linear recurrence relation ( $L_{n+2} = L_{n+1} + L_{n}$ ), it has a closed form expression $L_n = \left(\dfrac{\sqrt{5} + 1}{2}\right)^n + \left(\dfrac{1 - \sqrt{5}}{2}\right)^n$

We start with $f'(x) = \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} L_{n+1} x^n$ The expression for $L_n$ indicates that this is just a geometric sequence that can be written as $f'(x) = \sum_{n=0}^{\infty} \left(\dfrac{\sqrt{5} + 1}{2}\right) \left(\dfrac{\sqrt{5} + 1}{2} x\right)^n + \left(\dfrac{1 - \sqrt{5}}{2}\right)\left(\dfrac{1 - \sqrt{5}}{2} x\right)^n$

Now, $x \in \left[0, \dfrac{1}{2}\right]$ , so the sequence converges. Summing the sequence and integrating both sides from $0$ to $\dfrac{1}{2}$ gives us $f\left(\dfrac{1}{2}\right) - f(0) = \int_0^{\frac{1}{2}} \left(\dfrac{\sqrt5 + 1}{2 - x(\sqrt5 + 1)} + \dfrac{1 - \sqrt5}{2 - x(1 - \sqrt5)}\right) \text{d}x =\int_0^{\frac{1}{2}} \frac{2x+1}{1-x-x^2} \text{d}x$

Hence, $f(x) = -\ln(1-x-x^2)$ , by which $f\left(\dfrac{1}{2}\right) = \ln{4}$