Using the principle of mathematical induction to show that
`tan^(−1)(x/(1+1.2.x^2))+tan^(−1)(x/(1+2.3.x^2))+.....+tan^(−1)(x/(1+n(n+1)x^2)) =` `tan^(-1)(n+1)x-tan^(-1)x , forall x in N`.

Explanation

Let `P(n):tan^(-1)((x)/(1+1.2.x^2))+tan^(-1)((x)/(1+2.3.x^2))+.....+tan^(-1)((x)/(1+n(n+1)x^2))` .....(i)
`=tan^(-1)(n+1)x-tan^(-1)x`
Step I For `n=1`.
LHS of Eq. (i) `=tan^(-1)((x)/(1+1.2.x^2))`
`=tan^(-1)((2x-x)/(1+2x.x))=tan^(-1)2x-tan^(-1)x`
=RHS of Eq.(i)
Therefore , P(1) is true.
Step II Assume it is true for `n=k`.
`P(k):tan^(-1)((x)/(1+1.2x^2))+tan^(-1)((x)/(1+2.3x^2))+......+tan^(-1)((x)/([1+k(k+1)x^2]))`
`=tan^(-1)(k+1)x-tan^(-1)x`
Step III For `n=k+1`.
`P(k+1):tan^(-1)((x)/(1+1.2.x^2))+tan^(-1)((x)/(1+2.3.x^2))+.....+tan^(-1)((x)/(1+k(k+1)x^2))+......+tan^(-1)((x)/(1+(k=1)(k+2)x^2))`
`=tan^(-1)(k+2)x-tan^(-1)x`
`therefore LHS =tan^(-1)((x)/(1+1.2.x^2))`
`+tan^(-1)((x)/(1+2.3.x^2))+....+tan^(-1)((x)/(1+k(k+1)x^2))+tan^(-1)((x)/(1+(k+1)(k+2)x^2))`
`tan^(-1)(k+1)x-tan^(-1)x+tan^(-1)((x)/(1+(k+1)(k+2)x^2))`
`=tan^(-1)(k+1)x-tan^(-1)x+tan^(-1)(((k+2)x-(k+1)x)/(1+(k+2)x(k+1)x))`
`=tan^(-1)(k+1)x-tan^(-1)x+tan^(-1)(k+2)x-tan^(-1)(k+1)x=tan^(-1)(k+2)x-tan^(-1)x=RHS`
This shows that the result is true for `n=k+1`. Hence , by the principle of mathematical induction , the result is true for all the `n in N`.