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)`

Options

A. `tan^(-1) 1`
B. `tan^(-1) n`
C. `tan^(-1) (n + 1)`
D. `tan^(-1) (n -1)`

Answer

B

Explanation

`T_(n) = "sec"^(-1) sqrt(((n^(2) +1) (n^(2) -2n + 2))/((n^(2) -n +1)^(2)))`
`rArr sec^(2) T_(n) = ((n^(2) + 1) (n^(2) -2n + 2))/((n^(2) -n + 1)^(2))`
`rArr sec^(2) T_(n) = ((n^(2) +1)^(2) + (n^(2) +1) -2n(n^(2) + 1))/((n^(2) - n+ 1)^(2))`
`rArr sec^(2) T_(n) = (1 + (n^(2) + 1 -n)^(2))/((n^(2) -n + 1)^(2))`
`rArr tan T_(n) = (1)/(n^(2) -n + 1)`
`rArr tan T_(n) = (n -(n -1))/(1 + n(n -1)) = tan^(-1) n - tan^(-1) (n -1)`
`:. S = T_(1) + T_(2) + T_(3) +..+ T_(n)`
`:. S = tan^(-1) n`