A) \[{{\tan }^{-1}}(n+1)-\pi /4\]
B) \[\pi /4\]
C) \[{{\tan }^{-1}}(n+1)\]
D) 1
Correct Answer: A
Solution :
[a] Let \[\theta =\cos e{{c}^{-1}}\sqrt{({{n}^{2}}+1)({{n}^{2}}+2n+2)}\] \[\Rightarrow \cos e{{c}^{2}}\theta =({{n}^{2}}+1)({{n}^{2}}+2n+2)\] \[={{({{n}^{2}}+1)}^{2}}+2n({{n}^{2}}+1)+{{n}^{2}}+1\] \[={{({{n}^{2}}+n+1)}^{2}}+1\Rightarrow {{\cot }^{2}}\theta ={{({{n}^{2}}+n+1)}^{2}}\] \[\Rightarrow \tan \theta =\frac{1}{{{n}^{2}}+n+1}=\frac{(n+1)-n}{1+(n+1)n}\] \[\Rightarrow \theta ={{\tan }^{-1}}\left[ \frac{(n+1)-n}{1+(n+1)n} \right]={{\tan }^{-1}}(n+1)-ta{{n}^{-1}}n\] Thus, sum n terms of the given series \[=(ta{{n}^{-1}}2-ta{{n}^{-1}}1)+(ta{{n}^{-1}}3-ta{{n}^{-1}}2)\] \[+(ta{{n}^{-1}}4-ta{{n}^{-1}}3)+...+(ta{{n}^{-1}}(n+1)-ta{{n}^{-1}}n)\] \[\Rightarrow {{\tan }^{-1}}(n+1)-\pi /4\]
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