A random variable $X$ takes values $0,1,2,3$,... with probabilities. $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{2}\right)^x, \mathrm{k}$ is a constant, then $\mathrm{P}(\mathrm{X}=1)=$

Normalization yields the constant $k$. For a discrete random variable, the probabilities sum to 1: $\sum_{x=0}^{\infty} P(X=x) = 1$.

The infinite series $\sum_{x=0}^{\infty} (x+1)\left(\frac{1}{2}\right)^x$ is evaluated by differentiating the geometric series sum $\sum_{x=0}^{\infty} r^x = \frac{1}{1-r}$ with respect to $r$, yielding $\frac{d}{dr}\left(\frac{1}{1-r}\right) = \frac{1}{(1-r)^2}$.

Substituting $r = \frac{1}{2}$, the sum becomes $\frac{1}{\left(\frac{1}{2}\right)^2} = 4$. Then, $k \cdot 4 = 1$ implies $k = \frac{1}{4}$.

The desired probability is $P(X=1) = \frac{1}{4}(1+1)\left(\frac{1}{2}\right)^1 = \frac{1}{4}$.

This result corresponds to option C.