x[n] = \cos (2 \pi f_1 n) + \cos (2 \pi f_2 n) + w[n]

w[n] is zero-mean WGN with variance \sigma^2

I have to derive the expected value of this expression

 E \{\displaystyle{\frac{4 \pi^2}{\sigma^2} \sum_{n=0}^{N-1} n^2 (x[n] \cos (2 \pi f_1 n) - \cos (4 \pi f_1 n) - \cos (2 \pi f_1 n)\cos (2 \pi f_2 n))} \}

 = \displaystyle{\frac{4 \pi^2}{\sigma^2} \sum_{n=0}^{N-1} n^2 E \{x[n] \cos (2 \pi f_1 n) - \cos (4 \pi f_1 n) - \cos (2 \pi f_1 n)\cos (2 \pi f_2 n)} \}

 = \displaystyle{\frac{4 \pi^2}{\sigma^2} \sum_{n=0}^{N-1} n^2 E  \{ \cos^2 (2 \pi f_1 n) + \cos (2 \pi f_1 n) \cos (2 \pi f_2 n) + w[n] \cos(2 \pi f_1 n) - \cos (4 \pi f_1 n) - \cos (2 \pi f_1 n)\cos  (2 \pi f_2 n)} \}

 = \displaystyle{\frac{4 \pi^2}{\sigma^2} \sum_{n=0}^{N-1} n^2 E \{ \cos^2 (2 \pi f_1 n) + w[n] \cos(2 \pi f_1 n) - \cos (4 \pi f_1 n)} \}

 = \displaystyle{\frac{4 \pi^2}{\sigma^2} \sum_{n=0}^{N-1} n^2 E \{ \frac{1}{2} + \frac{\cos (4 \pi f_1 n)}{2} + w[n] \cos(2 \pi f_1 n) - \cos (4 \pi f_1 n)} \}

 = \displaystyle{\frac{4 \pi^2}{\sigma^2} \sum_{n=0}^{N-1} n^2 \frac{1}{2}} = \displaystyle{\frac{2 \pi^2}{\sigma^2} \sum_{n=0}^{N-1} n^2

could someone please check this?