I was reading the book "Famous Puzzles of Great Mathematicians" and in page 13 it states that:
Since the influence of the second term $((1-\sqrt{5})/2)^n/\sqrt{5}$ in the explicit formula for the nth term in fibonacci sequence $F_n=\frac{1}{\sqrt{5}}\left[(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n\right]$ is negligible because $|(1-\sqrt{5})/2|<1$, it is sufficient to calculate the first term $((1+\sqrt{5})/2)^n/\sqrt{5}$ and round off the result to the nearest integer to obtain the exact(integer) value of $F_n$. To be more exact, $$F_n=\Bigg\lfloor{\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n+\frac{1}{2}\right]}\Bigg\rfloor.$$
My problem is that I can't fill the gaps for example where does $+\frac{1}{2}$ come from? I saw this related question but the thing is that already assumes the solution and reverse engineering doesn't answer me.
That looks odd... As the comment says, the second term in Binet's formula is small and can be neglected, i.e.
$$ F_n \approx \frac1{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n $$
The estimate is good enough that it actually yields the Fibonacci numbers when rounded:
$$F_n = \left\lceil\frac1{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n\right\rfloor $$ and then express rounding to nearest by the floor function:
$$\lceil x \rfloor = \lfloor x + 0.5 \rfloor$$
so that
$$F_n = \left\lfloor\frac1{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n +\frac12\right\rfloor $$
To see why this works, take the 2nd term of Binet's formula and estimate it like
$$\left|\frac1{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n\right| = \frac1{\sqrt{5}}\left|\frac{1-\sqrt{5}}{2}\right|^n \leqslant \frac1{\sqrt{5}} < 0.5$$
The un-rounded values are:
$$\begin{align} F_0 &\approx 0.44 \to 0 \\ F_1 &\approx 0.72 \to 1\\ F_2 &\approx 1.17 \to 1\\ F_3 &\approx 1.89 \to 2\\ F_4 &\approx 3.07 \to 3\\ F_5 &\approx 4.96 \to 5\\ \end{align}$$ etc.
But your formula is adding $1/(2\sqrt5)$ which is odd...