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

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

I have to derive the expected value of this expression

$\displaystyle 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))} \} $

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

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

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

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

$\displaystyle = \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?