Saved Bookmarks
| 1. |
`sum_(n=1)^(prop)tan^(-1)((4n)/(n^(4)+5))=`A. `(pi)/(4)+tan^(-1)2`B. `(3pi)/(4)-tan^(-1)2`C. `tan^(-1)3`D. `(pi)/(4)+cot^(-1)2` |
|
Answer» Correct Answer - B::C::D `sum_(n=1)^(infty)tan^(-1)((4n)/(n^(4)+5))=sum_(n=1)^(infty)tan^(-1)((4n)/(1+(n^(2)+2)^(2)-4n^(2)))` `=sum_(n=1)^(infty)tan^(-1)((4n)/(1+(n^(2)+2n+2)(n^(2)-2n+2)))` `sum_(n=1)^(infty)[tan^(-1)(n^(2)+2n+2)-tan^(-1)(n^(2)-2n+2)]` `=2tan^(-1)(infty)-tan^(-1)1-tan^(-1)2=(3pi)/(4)-tan^(-1)2` `=(pi)/(4)+tan^(-1)(1)/(2)` |
|