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$\lim _{n \rightarrow \infty} \frac{\sqrt{n}}{\sqrt{\left(n^{3}\right)}}+\frac{\sqrt{n}}{\sqrt{(n+4)^{3}}}+\frac{\sqrt{n}}{\sqrt{(n+8)^{3}}}+\cdots \cdots \cdots+\frac{\sqrt{n}}{\sqrt{[n+4(n-1)]^{3}}}$ is
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The correct answers are:
$\frac{5-\sqrt{5}}{10}$
$\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{\sqrt{n}}{\sqrt{(n+4 r)^{3}}}$
$=\sum_{r=0}^{n-1} \frac{1}{n}\left(\frac{n \sqrt{n}}{\sqrt{(n+4 r)^{3}}}\right)$
$=\sum_{r=0}^{n-1} \frac{1}{n}\left(\frac{1}{\left(1+\frac{4 r}{n}\right)^{3 / 2}}\right)$
$=\int_{0}^{1} \frac{d x}{(1+4 x)^{3 / 2}}$
$=\frac{1}{4} \int_{1}^{5} \frac{d z}{z^{3 / 2}}=\left(1 / 4\left(\frac{-2}{\sqrt{z}}\right)\right)_{1}^{5}=\frac{5-\sqrt{5}}{10}$
$=\sum_{r=0}^{n-1} \frac{1}{n}\left(\frac{n \sqrt{n}}{\sqrt{(n+4 r)^{3}}}\right)$
$=\sum_{r=0}^{n-1} \frac{1}{n}\left(\frac{1}{\left(1+\frac{4 r}{n}\right)^{3 / 2}}\right)$
$=\int_{0}^{1} \frac{d x}{(1+4 x)^{3 / 2}}$
$=\frac{1}{4} \int_{1}^{5} \frac{d z}{z^{3 / 2}}=\left(1 / 4\left(\frac{-2}{\sqrt{z}}\right)\right)_{1}^{5}=\frac{5-\sqrt{5}}{10}$
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