InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The sum of series `sec^(-1)sqrt(2)+sec^(-1)(sqrt(10))/3+sec^(-1)(sqrt(50))/7++sec^(-1)sqrt(((n^2+1)(n^2-2n+2))/((n^2-n+1)^2))`is`tan^(-1)1`(b) `n``tan^(-1)(n+1)`(d) `tan^(-1)(n-1)` |
|
Answer» Let `S = sec^-1sqrt2+sec^-1(sqrt10/3)+sec^-1(sqrt50/7)+...+sec^-1sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2)` Here, `T_n = sec^-1sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2)` Let `sec^-1sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2) = theta` `=>sec theta =sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2` `=>sec^2theta = ((n^2+1)(n^2-2n+2))/(n^2-n+1)^2` `=>sec^2theta = ((n^2+1)(n^2+1-2n+1))/(n^2-n+1)^2` `=>1+tan^2theta = ((n^2+1)(n^2+1)-2n(n^2+1)+n^2+1)/(n^2-n+1)^2` `=>1+tan^2theta = ((n^2+1-n)^2+1)/(n^2-n+1)^2` `=>1+tan^2theta = 1+1/(n^2+1-n)^2` `=>tan^2theta = 1/(n^2+1-n)^2` `=>tan theta = 1/(1+n^2-n) = (n - (n-1))/(1+n(n-1))` `=>theta = tan^-1((n - (n-1))/(1+n(n-1)))` `=>theta = tan^-1(n) - tan^-1(n-1)` `:. T_n =tan^-1(n) - tan^-1(n-1)` Now, `S = tan^-1(1) - tan^-1(0) + tan^-1(2) - tan^-1(1)+tan^-1(3) - tan^-1(2)+...+tan^-1(n) - tan^-1(n-1)` `=>S = tan^-1(n) - tan^-1(0)` `=>S = tan^-1(n) - 0` `=>S = tan^-1(n)` |
|