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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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