If y = ⁻¹ ( 1 x^2+x+1 ) + ⁻¹ ( 1 x^2+3x+3 ) + ⁻¹ ( 1 x^2+5x+7 ) + upto n terms, then y'(0) =

Step-by-step solution

The given series can be written as the sum of n terms where the r -th term is given by: T_r = ⁻¹ ( 1 1 + (x+r-1)(x+r) ) Using the identity ⁻¹ ( A-B 1+AB ) = ⁻¹(A) - ⁻¹(B) , we can rewrite T_r as: T_r = ⁻¹ ( (x+r) - (x+r-1) 1 + (x+r)(x+r-1) ) = ⁻¹(x+r) - ⁻¹(x+r-1) Writing the first few terms: T₁ = ⁻¹(x+1) - ⁻¹(x) T₂ = ⁻¹(x+2) - ⁻¹(x+1) T₃ = ⁻¹(x+3) - ⁻¹(x+2) T_n = ⁻¹(x+n) - ⁻¹(x+n-1) Summing these n terms, all intermediate terms cancel out, leaving: y = _ r=1 ^ n T_r = ⁻¹(x+n) - ⁻¹(x) Differentiating with respect to