1. Let $g(x)=\sum_{i=1}^{n} \mathrm{C}_{n}^{i} \frac{i x^{i}(1-x)^{n-i}}{n}$. Then $g(2014)=$ $\qquad$
$$
-, 1.2014
$$

From $\frac{r}{n} \mathrm{C}_{n}^{r}=\mathrm{C}_{n-1}^{r-1}$, we get
$$
g(x)=x \sum_{k=0}^{n-1} \mathrm{C}_{n-1}^{k} x^{k}(1-x)^{n-k}=x \text {. }
$$

Thus, $g(2014)=2014$.