Plz solve this ....the question is from inverse trignometric series....thank u

denominator can be written as

1+(n-1)^2*(n+1)^2

numerator can be written as

(n+1)^2 – (n – 1)^2

since arctan[(a – b)/(1+ab)]= arctana – arctanb,

so we can write the nth term as

arctan(n+1)^2 – arctan(n – 1)^2, and this is obviously easy to calculate, since the sum TELESCOPES. if you dont know telescopic series method, you can google it as it is too long to explain here.