A) \[\frac{20}{401+20\operatorname{n}}\]
B) \[\frac{\operatorname{n}}{{{\operatorname{n}}^{2}}+20\operatorname{n}+1}\]
C) \[\frac{20}{{{\operatorname{n}}^{2}}+20\operatorname{n}+1}\]
D) \[\frac{\operatorname{n}}{401+20\operatorname{n}}\]
Correct Answer: C
Solution :
We know that, \[{{\tan }^{-1}}\frac{1}{1+2}+{{\tan }^{-1}}\frac{1}{1+2\times 3}+{{\tan }^{-1}}\frac{1}{1+3\times 4}+......+\]\[{{\tan }^{-1}}\frac{1}{1+(n-1)n}+{{\tan }^{-1}}\frac{1}{1+n(n+1)}+......+\] \[{{\tan }^{-1}}\frac{1}{1+(n+19)(n+20)}={{\tan }^{-1}}\frac{n+19}{n+21}\] \[\Rightarrow \]\[{{\tan }^{-1}}\frac{n-1}{n+1}+{{\tan }^{-1}}\frac{1}{1+n(n+1)}+{{\tan }^{-1}}\frac{1}{1+(n+1)(n+2)}\]\[+.....+\frac{1}{1+(n+19)(n+20)}={{\tan }^{-1}}\frac{n+19}{n+21}\] \[{{\tan }^{-1}}\frac{1}{1+n(n+1)}={{\tan }^{-1}}\frac{1}{1+(n+1)(n+2)}+......+\]\[\frac{1}{1+(n+19)(n+20)}={{\tan }^{-1}}\frac{n+19}{n+21}-{{\tan }^{-1}}\frac{n-1}{n+1}\] \[{{\tan }^{-1}}\left( \frac{1}{{{n}^{2}}+n+1} \right)+{{\tan }^{-1}}\left( \frac{1}{{{n}^{2}}+3n+3} \right)+.....+\] \[{{\tan }^{-1}}\frac{1}{1+(n+19)(n+20)}\] \[={{\tan }^{-1}}\left( \frac{\frac{n+19}{n+21}+\frac{n-1}{n+1}}{1+\frac{n+19}{n+21}\times \frac{n-1}{n+1}} \right)\] \[={{\tan }^{-1}}\frac{20}{{{n}^{2}}+20n+1}=S\] \[\therefore \]\[{{\tan }^{-1}}S=\frac{20}{{{n}^{2}}+20n+1}\]
You need to login to perform this action.
You will be redirected in
3 sec
