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A) \[\dfrac{1}{35}\] B) \[\dfrac{1}{14}\] C) \[\dfrac{1}{10}\] D) \[\dfrac{1}{5}\]

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Hint: Try to convert the sum into integral using the idea that integral is a limit of a sum.

We can see that the \[{{r}^{th}}\] term of the sum is written as \[\dfrac{\sqrt{n}}{\sqrt{r}{{(3\sqrt{r}+4\sqrt{n})}^{2}}}\]

Therefore, multiplying by \[\dfrac{1}{\sqrt{n}}\]from both numerator and denominator we rewrite our equation as,

= \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{4n}{\dfrac{1}{\sqrt{\dfrac{r}{n}}{{(3\sqrt{\dfrac{r}{n}}+4)}^{2}}}}\dfrac{1}{n}\]

This limit can now be replaced by an integral as \[\dfrac{1}{n}=dx\]and \[\dfrac{k}{n}=x\].

The upper and lower limits of the integral can be found by evaluating the limit of \[\dfrac{k}{n}\]as \[k\] tends to upper and lower bounds of our summation, which in our case are \[0\] (as lower bound) and \[1\]as upper bound. Every limit of the form

\[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=an}^{bn}{\dfrac{1}{n}f\left( \dfrac{k}{n} \right)}\] can be converted to the integral \[\int\limits_{a}^{b}{f(x)}dx\]. The summation is known as the Reimann sum of the integral. It helps to convert such infinite sums into an integral.

So,

= \[\int\limits_{0}^{1}{\dfrac{1}{\sqrt{x}{{\left( 3\sqrt{x}+4 \right)}^{2}}}}dx\]

For evaluating this integral substitute\[\sqrt{x}=t\]. So\[dt=\dfrac{1}{2\sqrt{x}}dx\]. Upper bound becomes \[1\] and lower bound becomes \[0\](which remain the same in our case as the substitution was \[\sqrt{x}\] as\[t\], but may change if it would have been something else)

= \[\int\limits_{0}^{1}{\dfrac{2}{{{\left( 3t+4 \right)}^{2}}}}dt\]

= \[\left[ \dfrac{-2}{3\left( 3t+4 \right)} \right]_{0}^{1}\]

= \[\left( \dfrac{-2}{21} \right)-\left( \dfrac{-2}{12} \right)\]

=\[\dfrac{1}{14}\]

So, the value of the limit is option B) \[\dfrac{1}{14}\]

Note: Students have to be careful when writing the \[{{k}^{th}}\]term of the summation and its limits. Also, when converting the limit to the integral one should be careful. Students can use their own methods i.e. any other substitution when solving the integral. They may use either the lower Reimann sum or the upper Reimann sum, in both cases the answer will be the same.

We can see that the \[{{r}^{th}}\] term of the sum is written as \[\dfrac{\sqrt{n}}{\sqrt{r}{{(3\sqrt{r}+4\sqrt{n})}^{2}}}\]

Therefore, multiplying by \[\dfrac{1}{\sqrt{n}}\]from both numerator and denominator we rewrite our equation as,

= \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{4n}{\dfrac{1}{\sqrt{\dfrac{r}{n}}{{(3\sqrt{\dfrac{r}{n}}+4)}^{2}}}}\dfrac{1}{n}\]

This limit can now be replaced by an integral as \[\dfrac{1}{n}=dx\]and \[\dfrac{k}{n}=x\].

The upper and lower limits of the integral can be found by evaluating the limit of \[\dfrac{k}{n}\]as \[k\] tends to upper and lower bounds of our summation, which in our case are \[0\] (as lower bound) and \[1\]as upper bound. Every limit of the form

\[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=an}^{bn}{\dfrac{1}{n}f\left( \dfrac{k}{n} \right)}\] can be converted to the integral \[\int\limits_{a}^{b}{f(x)}dx\]. The summation is known as the Reimann sum of the integral. It helps to convert such infinite sums into an integral.

So,

= \[\int\limits_{0}^{1}{\dfrac{1}{\sqrt{x}{{\left( 3\sqrt{x}+4 \right)}^{2}}}}dx\]

For evaluating this integral substitute\[\sqrt{x}=t\]. So\[dt=\dfrac{1}{2\sqrt{x}}dx\]. Upper bound becomes \[1\] and lower bound becomes \[0\](which remain the same in our case as the substitution was \[\sqrt{x}\] as\[t\], but may change if it would have been something else)

= \[\int\limits_{0}^{1}{\dfrac{2}{{{\left( 3t+4 \right)}^{2}}}}dt\]

= \[\left[ \dfrac{-2}{3\left( 3t+4 \right)} \right]_{0}^{1}\]

= \[\left( \dfrac{-2}{21} \right)-\left( \dfrac{-2}{12} \right)\]

=\[\dfrac{1}{14}\]

So, the value of the limit is option B) \[\dfrac{1}{14}\]

Note: Students have to be careful when writing the \[{{k}^{th}}\]term of the summation and its limits. Also, when converting the limit to the integral one should be careful. Students can use their own methods i.e. any other substitution when solving the integral. They may use either the lower Reimann sum or the upper Reimann sum, in both cases the answer will be the same.

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