71Seems like I have seen this question on TIIT only, some before.
\hspace{-16}$If $\bf{\mathbb{A}=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}}$\\\\\\ Then $\bf{\lfloor \mathbb{A}\rfloor =}$
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71
341We can first prove that
\frac{1}{(k+1)\sqrt k} < 2 \left(\frac{1}{\sqrt k} - \frac{1}{\sqrt{k+1}} \right).
i.e. that \frac{1}{\sqrt{k+1}} < \frac{2}{\sqrt k + \sqrt{k+1}} \right) or that \sqrt{k+1} > \sqrt k which is obvious.
Hence the sum can be bounded by
2 \left(1 - \frac{1}{10} \right)<2
For k≥4, we have
\frac{1}{(k+1)\sqrt k} \ge \frac{2}{(k+1)k} = 2 \left(\frac{1}{k} - \frac{1}{k+1} \right)
Hence the given sum is greater than
\frac{1}{2} + \frac{1}{3 \sqrt 2} + 2 \left(\frac{1}{4} - \frac{1}{100} \right)>\frac{1}{2} + \frac{1}{2} + \left(\frac{1}{9} - \frac{1}{50} \right)>1
Hence [A]=1
