If ${ }^{\mathrm{n}} \mathrm{C}_0+\frac{1}{2}{ }^{\mathrm{n}} \mathrm{C}_1+\frac{1}{3}{ }^{\mathrm{n}} \mathrm{C}_2+\ldots \ldots \ldots \ldots . . \cdot \frac{1}{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-1}+\frac{1}{\mathrm{n}+1}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}=\frac{1023}{10}$ then $\mathrm{n}=$

Given the sum $S = \binom{n}{0} + \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \ldots + \frac{1}{n+1}\binom{n}{n}$, the general term is $\frac{1}{r+1}\binom{n}{r}$.

Using the identity $\frac{1}{r+1}\binom{n}{r} = \frac{1}{n+1}\binom{n+1}{r+1}$, the sum becomes $S = \frac{1}{n+1} \sum_{r=0}^{n} \binom{n+1}{r+1}$.

Substituting $k = r+1$ yields $S = \frac{1}{n+1} \sum_{k=1}^{n+1} \binom{n+1}{k}$.

From the binomial theorem, $\sum_{k=0}^{n+1} \binom{n+1}{k} = 2^{n+1}$, so $\sum_{k=1}^{n+1} \binom{n+1}{k} = 2^{n+1} - 1$.

Thus, $S = \frac{2^{n+1} - 1}{n+1}$.

Setting this equal to $\frac{1023}{10}$ gives $\frac{2^{n+1} - 1}{n+1} = \frac{1023}{10}$.

Testing $n=9$ yields $\frac{2^{10} - 1}{10} = \frac{1023}{10}$, matching the given value.

The correct value is $n=9$, corresponding to option C.