(²ⁿ⁺¹C₀)² - (²ⁿ⁺¹C₁)² + (²ⁿ⁺¹C₂) – …... – (²ⁿ⁺¹C_2n+1) = ?

Please answer the above series

Seeing the question it is clear that ans will be of the form (a-b) ^2n+1 Expanding (a-b) ^2n+1a^2n+1- 2n+1(a^2n)b + n ( 2n+1)a^2n-1 b^2_ _ _ _ - b^2n+1 __1Now from(1) and the given equation a^2n+1=1 __22n+1(a^2n)b=(2n+1)^2= a^2n b = 2n +1 __3n(2n+1)a^2n-1 b^2 = n(2n +1)= a^2n-1 b^2 =1 __4b^2n+1 =1 __5Now dividing 4/3a^2n b/a^2n-1 b^2 = 2n + 1= a/b = 2n + 1 __6Now solving (2)&(5)a^2n+1 = b^2n+1=(a/b)^2n+1 = 1=(2n+1)^2n+1 = 1 (by 6)Let 2n+1 = x __7Then, x^x = 1= x=0, or x= 1By 72n+1 = 0, or 2n+1 =1Now as 2n+1 is odd, it is never equal to 0=> 2n+1 =1 __8Using the value in equation (2)&(5),We get, a=1 & b=1Now, (a-b) ^2n+1 = (1-1)^2(0)+1=0^1 =0 Answer = 0