As series resistors are added, the resistance is added directly so the total resistance will be equal to
[tex]R_T = R_1+R_2+R_3... R_{\infty}[/tex]
Since the power is determined as
[tex]P = \frac{V^2}{R}[/tex]
We have there that
[tex]P \propto \frac{1}{R}[/tex]
The power is inversely proportional to the increase in resistance, so it will tend to decrease as more resistors are added in series.
The correct answer is: C.