The value of mathop {text{Lim}}limits_{n to i | vedclass

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(C) Let the given sum be $S_n = \sum_{r=1}^{4n} \frac{\sqrt{n}}{\sqrt{r}(3\sqrt{r} + 4\sqrt{n})^2}$.Dividing numerator and denominator by $n\sqrt{n}$,

Vedclass · Accepted Answer

$\frac{1}{{10}}$

Vedclass · Answer

(C) Let the given sum be $S_n = \sum_{r=1}^{4n} \frac{\sqrt{n}}{\sqrt{r}(3\sqrt{r} + 4\sqrt{n})^2}$.Dividing numerator and denominator by $n\sqrt{n}$,we get:$S_n = \frac{1}{n} \sum_{r=1}^{4n} \frac{1}{\sqrt{\frac{r}{n}} (3\sqrt{\frac{r}{n}} + 4)^2}$.As $n 	o \infty$,this sum becomes the definite integral:$S = \int_0^4 \frac{dx}{\sqrt{x}(3\sqrt{x} + 4)^2}$.Let $t = 3\sqrt{x} + 4$. Then $dt = \frac{3}{2\sqrt{x}} dx$,which implies $\frac{dx}{\sqrt{x}} = \frac{2}{3} dt$.When $x = 0$,$t = 4$. When $x = 4$,$t = 3(2) + 4 = 10$.Substituting these into the integral:$S = \int_4^{10} \frac{1}{t^2} \cdot \frac{2}{3} dt = \frac{2}{3} \left[ -\frac{1}{t} ight]_4^{10} = \frac{2}{3} \left( -\frac{1}{10} + \frac{1}{4} ight) = \frac{2}{3} \left( \frac{-2 + 5}{20} ight) = \frac{2}{3} \cdot \frac{3}{20} = \frac{1}{10}$.

The value of $\mathop {\text{Lim}}\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^{4n} {\frac{{\sqrt n }}{{\sqrt r {{\left( {\,3\sqrt r + 4\sqrt n \,} \right)}^2}}}} $ is equal to

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