Find the value of `^4n C_0+^(4n)C_4+^(4n)C_8++""^(4n)C_(4n)` .

Explanation

`because 4 - 0 = 8 - 4 = …= 4 `
` therefore ` Four roots of unity `(1)^(1/4) ` are 1, - 1, -I , we have
` (1 + x)^(4n) = ""^(4n)C_(0) + ""^(4n)C_(1) x + ""^(4nC_(4) x^(2) + ""^(4n)C_(3)x^(3) + …`
Putting x = 1 , I , - I and then adding , we get
` 4 (""^(4n)C_(0) + ""^(4n)C_(4) + ""^(4n)C_(8) + ...) = 2^(4n) + 0 + (1 + i)^(4n) + (1 - i)^(4n) `
`= 3^(4n) + (2i)^(2n) + (-2i)^(2n)`
` 2^(4n) + 2^(2n) (-1)^(n) + 2^(2n) (-1)^(n)`
` = 2^(4n) + (-1)^(n)* 2^(2n +1)`
` therefore ""^(2n)C_(0) + ""^(4n)C_(4) + ""^(4n)C_(8) + ...= 2^(4n-2) + (-1)^(n) * 2^(2n-1)`