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.
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by
Tapas Mazumdar