If frac{1}{sqrt{1}+sqrt{2}}+frac{1}{sqrt{2}+s | vedclass

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(D) For $m$: $\frac{1}{\sqrt{k}+\sqrt{k+1}} = \frac{\sqrt{k+1}-\sqrt{k}}{k+1-k} = \sqrt{k+1}-\sqrt{k}$.Summing from $k=1$ to $99$: $(\sqrt{2}-\sqrt{1}

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$11 x-100 y=0$

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(D) For $m$: $\frac{1}{\sqrt{k}+\sqrt{k+1}} = \frac{\sqrt{k+1}-\sqrt{k}}{k+1-k} = \sqrt{k+1}-\sqrt{k}$.Summing from $k=1$ to $99$: $(\sqrt{2}-\sqrt{1}) + (\sqrt{3}-\sqrt{2}) + \ldots + (\sqrt{100}-\sqrt{99}) = \sqrt{100}-\sqrt{1} = 10-1 = 9$.So,$m=9$.For $n$: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.Summing from $k=1$ to $99$: $(1-\frac{1}{2}) + (\frac{1}{2}-\frac{1}{3}) + \ldots + (\frac{1}{99}-\frac{1}{100}) = 1-\frac{1}{100} = \frac{99}{100}$.So,$n=\frac{99}{100}$.The point is $(m, n) = (9, \frac{99}{100})$.Checking the options,for option $D$: $11(9) - 100(\frac{99}{100}) = 99 - 99 = 0$.Thus,the point lies on the line $11x-100y=0$.

If $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$ and $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n$,then the point $(m, n)$ lies on the line

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