Answer
Verified
425.7k+ views
Hint: Try to convert the sum into integral using the idea that integral is a limit of a sum.
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.
Complete answer:
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 \[\dfrac{r}{n}=x\].
The upper and lower limits of the integral can be found by evaluating the limit of \[\dfrac{r}{n}\] as \[r\] 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
\[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=an}^{bn}{\dfrac{1}{n}f\left( \dfrac{r}{n} \right)}\] 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) \[\dfrac{1}{14}\]
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Two charges are placed at a certain distance apart class 12 physics CBSE
Difference Between Plant Cell and Animal Cell
What organs are located on the left side of your body class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The planet nearest to earth is A Mercury B Venus C class 6 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE