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$\frac{1}{\sqrt{100}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{98}}+\frac{1}{\sqrt{98}+\sqrt{97}}+\cdots+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{1}}$

If the above expression equals $k$, find $k^{2}$.

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