# Let's do some calculus! (26)

Calculus Level 4

$\large \displaystyle \left \lfloor \lim_{n \to \infty} { n^{{-3}/{2}} \sum_{j=1}^{6n} {\sqrt{j}}} \right \rfloor = \, ?$

Notation: $\lfloor \cdot \rfloor$ denotes the floor function.