B.E.

( 1 + x ) ⁿ = 1 + ⁿ C ₁ x + ⁿ C ₂ x ² + ................+ ⁿ C _n x ⁿ

Differentiating with respect to " x " ,

n ( 1 + x ) ^{n - 1} = ⁿ C ₁ + 2 ⁿ C ₂ x + ................ + n ⁿ C _n x ^{n - 1} ....................... ( 1 )

Now ,
1 + e ^{i θ} = 1 + cos θ + i sin θ = 2 cos ² θ + 2 i sin θ2 cos θ2
= 2 cos θ2 ( cos θ2 + i sin θ2 ) = 2 cos θ2 e ^{i θ2}

From ( 1 ) , putting " x = e ^{i θ} " and equating real parts ,

ⁿ C ₁ + 2 ⁿ C ₂ cos θ + ................ + n ⁿ C _n cos ( n - 1 ) θ = 2 ⁿ cos ⁿ θ2 cos n θ2

The answer is wrong.The answer I have is n.2^n-1cos^n-1@2.cos(n-1)@2.

Yup , if you put " e ^{i θ} " in ( 1 ) , then you get the result you stated . I actually put " e ^{i θ} " in " ( 1 + x ) ⁿ " .

Thank you very much!