Using mathematical induction , prove that `tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+.....+tan^(-1)((1)/(n^2+n+1))=tan^(-1)((n)/(n+2))`

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `(2pi)/(3)`
D. 0

Explanation

We have
`tan^(-1)1/3+tan^(-1)1/7+tan^(-1)1/13+..+tan^(-1)(1)/(n^(2)+n_+1)+…to infty`
`=underset(nrarrinfty)lim underset(r=1)overset(n)Sigma tan^(-1){(1)/(1+r(r+1))}`
`=underset(nrarrinfty)lim underset(r=1)oveset(n)Sigma tan^(-1){(r+1)-r)/(1+r(r+1))`
`=underset(nrarrinfty)lim (tan^(-1)(n+1)-tan^(-1)1)`
`=tan^(-1)infty-tan^(-1)=(pi)/(12)-(pi)/(4)-(pi)/(4)`