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27/12/2021 12:26 pm
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The value of \(\lim\limits_{n \to \infty}\)\(\frac{1}{n}\) \(\displaystyle\sum_{r=0}^{2n-1}\) \(\frac{n^2}{n^2 + 4r^2}\) is
(a) \(\frac{1}{2}tan^{-1}\)4
(b) \(\frac{1}{4}tan^{-1}\)3
(c) \(\frac{1}{3}tan^{-1}\)4
(d) \(\frac{1}{5}tan^{-1}\)4
1 Answer
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27/12/2021 12:38 pm
Correct answer: (a) \(\frac{1}{2}tan^{-1}\)4
Explanation:
\(\lim\limits_{n \to \infty}\)\(\frac{1}{n}\) \(\displaystyle\sum_{r=0}^{2n-1}\) \(\frac{n^2}{n^2 + 4r^2}\)
= \(\int_0^2\)\(\frac{dx}{1+4x^2}\)
= \(\frac{1}{4}\)\(\int_0^2\)\(\frac{dx}{(\frac{1}{2})^2 + x^2}\)
= \(\Big[\frac{1}{2}tan^{-1}(2x) \Big]_0^2\)
= \(\frac{1}{2}tan^{-1}\)4