Let I_n = ₀^ /2 1 - 2nx 1 - 2x dx for n N . Then the sequence I₁, I₂, I₃, forms

Step-by-step solution

Given I_n = ₀^ /2 1 - 2nx 1 - 2x dx Using the identity 1 - 2 = 2 ^2 , we can write: I_n = ₀^ /2 2 ^2 nx 2 ^2 x dx = ₀^ /2 ^2 nx ^2 x dx Now, consider the difference between consecutive terms: I_ n+1 - I_n = ₀^ /2 ^2(n+1)x - ^2 nx ^2 x dx Using the identity ^2 A - ^2 B = (A+B) (A-B) , we get: I_ n+1 - I_n = ₀^ /2 (2n+1)x x ^2 x dx = ₀^ /2 (2n+1)x x dx Let K_ 2n+1 = ₀^ /2 (2n+1)x x dx . We find the difference of this new sequence: K_ 2n+1 - K_ 2n-1 = ₀^ /2 (2n+1)x - (2n-1)x x dx = ₀^ /2 2 2nx x x dx = 2 ₀^ /2 2nx dx