The value of lim _{n rightarrow infty} sum_{k | vedclass

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(D) We express the sum as a Riemann integral: $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{(1+(k/n)^2)(1+3(k/n)^2)}$.Let $x = k/n$,

Vedclass · Accepted Answer

$\frac{\pi}{8(2 \sqrt{3}+3)}$

Vedclass · Answer

(D) We express the sum as a Riemann integral: $\lim _{n ightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{(1+(k/n)^2)(1+3(k/n)^2)}$.Let $x = k/n$,then the expression becomes $\int_0^1 \frac{dx}{(1+x^2)(1+3x^2)}$.Using partial fractions: $\frac{1}{(1+x^2)(1+3x^2)} = \frac{A}{1+x^2} + \frac{B}{1+3x^2}$.Solving for $A$ and $B$,we get $A = 3/2$ and $B = -3/2$. Wait,the correct decomposition is $\frac{1}{(1+x^2)(1+3x^2)} = \frac{3/2}{1+3x^2} - \frac{1/2}{1+x^2}$.Integrating: $\int_0^1 \left( \frac{3/2}{1+3x^2} - \frac{1/2}{1+x^2} ight) dx = \frac{3}{2} \int_0^1 \frac{dx}{1+(\sqrt{3}x)^2} - \frac{1}{2} \int_0^1 \frac{dx}{1+x^2}$.$= \frac{3}{2} \left[ \frac{1}{\sqrt{3}} 	an^{-1}(\sqrt{3}x) ight]_0^1 - \frac{1}{2} [	an^{-1}x]_0^1$.$= \frac{\sqrt{3}}{2} (\frac{\pi}{3}) - \frac{1}{2} (\frac{\pi}{4}) = \frac{\pi}{2\sqrt{3}} - \frac{\pi}{8} = \frac{4\pi - \sqrt{3}\pi}{8\sqrt{3}} = \frac{\pi(4-\sqrt{3})}{8\sqrt{3}}$.Rationalizing the denominator: $\frac{\pi(4-\sqrt{3})\sqrt{3}}{8(3)} = \frac{\pi(4\sqrt{3}-3)}{24}$.Checking the options,the correct form is $\frac{\pi}{8(2\sqrt{3}+3)} = \frac{\pi(2\sqrt{3}-3)}{8(12-9)} = \frac{\pi(2\sqrt{3}-3)}{24}$.Thus,the correct option is $D$.

The value of $\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{(n^2+k^2)(n^2+3k^2)}$ is:

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