Correct choice is (b) \(\sum_{k=-\infty}^{\infty}|c_k|\)
Explanation: The average power of a periodic signal x(t) is given as \(\frac{1}{T_p}\int_{t_0}^{t_0+T_p}|x(t)|^2 dt\)
=\(\frac{1}{T_p}\int_{t_0}^{t_0+T_p} x(t).x^* (t) dt\)
=\(\frac{1}{T_p}\int_{t_0}^{t_0+T_p}x(t).\sum_{k=-∞}^∞ c_k^* e^{-j2πkF_0 t} dt\)
By interchanging the positions of integral and summation and by applying the integration, we get
=\(\sum_{k=-∞}^∞|c_k |^2\)