A) \[\frac{1}{n-1}\]
B) \[\frac{1}{n+1}\]
C) \[\frac{1}{2n-1}\]
D) \[\frac{1}{2n+1}\]
Correct Answer: A
Solution :
[a] Given: \[{{u}_{n}}=\int\limits_{0}^{\pi /4}{{{\tan }^{n}}\theta d\theta }\] \[=\int\limits_{0}^{\pi /4}{{{\tan }^{2}}\theta {{\tan }^{n-2}}\theta d\theta }\] \[=\int\limits_{0}^{\pi /4}{(se{{c}^{2}}\theta -1)ta{{n}^{n-2}}\theta d\theta }\] \[=\int\limits_{0}^{\pi /4}{se{{c}^{2}}\theta {{\tan }^{n-2}}\theta d\theta -\int\limits_{0}^{\pi /4}{{{\tan }^{n-2}}\theta d\theta }}\] \[=\int\limits_{0}^{\pi /4}{se{{c}^{2}}\theta {{\tan }^{n-2}}\theta d\theta -{{u}_{n-2}}}\] \[\Rightarrow {{u}_{n}}+{{u}_{n-2}}=\int\limits_{0}^{\pi /4}{{{\sec }^{2}}\theta {{\tan }^{n-2}}\theta d\theta }\] \[=\left. \frac{{{\tan }^{n-1}}\theta }{n-1} \right|_{0}^{\pi /4}=\frac{1}{n-1}\]
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