It is linear homogeneous recurrence, and its characteristic equation is given by
$r^{2} - r - 1 = 0$
Solving using quadratic formula we get,
$r_{1} = \dfrac{1+\sqrt{5}}{2}$ and $r_{2} = \dfrac{1-\sqrt{5}}{2}$
By "Distinct Roots Theorem"
$f_{n} = \alpha_{1}\bigg(\dfrac{(1 + \sqrt{5})}{2}\bigg)^{n} + \alpha_{2}\bigg(\dfrac{(1 - \sqrt{5})}{2}\bigg)^{n}$
Now we should find $\alpha_{1}$ and $\alpha_{2}$ using initial conditions.
$f_{0} = \alpha_{1} + \alpha_{2} = 0$
$f_{1} = \alpha_{1}\bigg(\dfrac{(1 + \sqrt{5})}{2} \bigg)+ \alpha_{2}\bigg(\dfrac{(1 - \sqrt{5})}{2}\bigg) = 1$
So, $\alpha_{1} = \dfrac{1}{\sqrt{5}}$ and $\alpha_{2} = \dfrac{-1}{\sqrt{5}}$
$f_{n} =\dfrac{1}{\sqrt{5}}\bigg(\dfrac{(1 + \sqrt{5})}{2}\bigg)^{n} + \dfrac{-1}{\sqrt{5}}\bigg(\dfrac{(1 - \sqrt{5})}{2}\bigg)^{n}$ is the solution.
So, the correct answer is $(B)$.
