[Number Theory]
Problem: Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. For example, $\lfloor 3.1\rfloor=3$ and $\lfloor-1.4\rfloor=-2$.

Suppose that $f(n)=2 n-\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ and $g(n)=2 n+\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ for each positive integer $n$.
Determine the value of $g(2011)$.

Solution: If $n=2011$, then $8 n-7=16081$ and so $\sqrt{8 n-7} \approx 126.81$.

Thus, $\frac{1+\sqrt{8 n-7}}{2} \approx \frac{1+126.81}{2} \approx 63.9$.

Therefore, $g(2011)=2(2011)+\left\lfloor\frac{1+\sqrt{8(2011)-7}}{2}\right\rfloor=4022+\lfloor 63.9\rfloor=4022+63=4085$.

Final answer: 4085