If a₁, a₂, a₃, , a_n are in arithmetic progression with common difference d, then [ ⁻¹ ( d 1 + a₁ a₂ ) + ⁻¹ ( d 1 + a₂ a₃ ) + + ⁻¹ ( d 1 + a_ n-1 a_n ) ] = ____

Step-by-step solution

Given a₁, a₂, , a_n are in arithmetic progression with common difference d . d = a₂ - a₁ = a₃ - a₂ = = a_n - a_ n-1 The general term of the series is: ⁻¹ ( d 1 + a_k a_ k+1 ) = ⁻¹ ( a_ k+1 - a_k 1 + a_k a_ k+1 ) Using the identity ⁻¹ ( x - y 1 + xy ) = ⁻¹ x - ⁻¹ y , we get: ⁻¹ ( a_ k+1 - a_k 1 + a_k a_ k+1 ) = ⁻¹ a_ k+1 - ⁻¹ a_k Let S be the sum of the inverse trigonometric series: S = _ k=1 ^ n-1 ⁻¹ ( d 1 + a_k a_ k+1 ) S = ( ⁻¹ a₂ - ⁻¹ a₁) + ( ⁻¹ a₃ - ⁻¹ a₂) + + ( ⁻¹ a_n - ⁻¹ a_ n-1 ) This is a telescoping sum, w