A) \[{{\tan }^{-1}}\left( \frac{n}{n+2} \right)\]
B) \[{{\cot }^{-1}}\left( \frac{n+2}{n} \right)\]
C) \[{{\tan }^{-1}}(n+1)-{{\tan }^{-1}}1\]
D) All of these
Correct Answer: D
Solution :
Let\[S=3+7+13+21+........+{{T}_{n}}\]\[\Rightarrow \]\[{{T}_{n}}={{n}^{2}}+n+1\]. Let \[{{T}_{r}}={{\cot }^{-1}}({{r}^{2}}+r+1)={{\tan }^{-1}}(r+1)-{{\tan }^{-1}}r\]. Put \[r=1,\ 2,.........,n\] and add, we get the required sum \[{{\tan }^{-1}}(n+1)-{{\tan }^{-1}}1={{\tan }^{-1}}\left( \frac{n}{n+2} \right)={{\cot }^{-1}}\left( \frac{n+2}{n} \right)\].
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