⁻¹ ( 1 1 + 1 2 ) + ⁻¹ ( 1 1 + 2 3 ) + + ⁻¹ ( 1 1 + n(n+1) ) =

Step-by-step solution

The general term of the given series is T_k = ⁻¹ ( 1 1 + k(k+1) ) . This can be written as T_k = ⁻¹ ( (k+1) - k 1 + (k+1)k ) . Using the identity ⁻¹ ( x - y 1 + xy ) = ⁻¹x - ⁻¹y , we get: T_k = ⁻¹(k+1) - ⁻¹k Writing the terms for k = 1, 2, , n : T₁ = ⁻¹2 - ⁻¹1 T₂ = ⁻¹3 - ⁻¹2 T_n = ⁻¹(n+1) - ⁻¹n Adding all the terms, the intermediate terms cancel out: S = _ k=1 ^ n T_k = ⁻¹(n+1) - ⁻¹1 Applying the inverse tangent difference formula again: S = ⁻¹ ( (n+1) - 1 1 + (n+1)(1) ) S = ⁻¹ ( n n+2 ) Answer: ⁻¹ ( n n+2 )