Mathematics

The value of $$ \displaystyle \lim_{n\rightarrow \infty } \sum_{r=1}^{4n} \dfrac{\sqrt{n}}{\sqrt{r}(3\sqrt{r}+4\sqrt{n})^{2}}$$ is equal to


ANSWER

$$\displaystyle \dfrac{1}{10}$$


SOLUTION
$$ \displaystyle \lim_{n\rightarrow \infty }\sum_{r=1}^{4n}\displaystyle \frac{\sqrt{n}}{\sqrt{r}\left (3\sqrt{r}+4\sqrt{n}  
\right )^{2}}$$

$$ \displaystyle T_{r}=\displaystyle \frac{1}{\sqrt{\frac{r}{n}}n\left (3\sqrt{\frac{r}{n}}+4  \right )^{2}}$$

$$ \displaystyle \Rightarrow S=\lim _{ n\rightarrow \infty  } \displaystyle \frac { 1 }{ n } \sum _{ r=1 }^{ 4n } \frac { 1 }{ \sqrt { \frac { r }{ n }  } { (3\sqrt { \frac { r }{ n }  } +4) }^{ 2 } } $$
Put $$\sqrt { \displaystyle \frac { r }{ n }  } =x$$

$$ \displaystyle \Rightarrow S=\int_{0}^{4}\displaystyle \frac{\mathrm{d} x}{\sqrt{x}\left (3\sqrt{x}+4  \right )^{2}}  $$
 
Put  $$3\sqrt{x}+4=t$$
$$ \Rightarrow \displaystyle  \frac{3}{2}.\frac{1}{\sqrt{x}}\mathrm{d} x=\mathrm{d} t$$

$$\displaystyle S= \frac{2}{3}\int_{4}^{10}\frac{\mathrm{d} t}{t^{2}}$$

$$=\displaystyle \frac{2}{3}\left [\frac{-1}{t}  \right ]_{4}^{10}$$

$$=\displaystyle \frac{2}{3}\left [\frac{1}{t}  \right ]_{10}^{4}$$

$$\displaystyle =\frac{1}{10}$$
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Single Correct Hard Published on 17th 09, 2020
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