If $$\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}$$ then $n$ is equal to :

$$\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}$$ <br/><br/>$$ \begin{aligned} & \Rightarrow \sum_{r=0}^n \frac{1}{r+1}{ }^n C_r=\frac{1023}{10} \\\\ & \quad\left( \because{ }^{n+1} C_{r+1}=\frac{n+1}{r+1}{ }^n C_r\right) \\\\ & \Rightarrow \sum_{r=0}^n \frac{1}{n+1}{ }^{n+1} C_{r+1}=\frac{1023}{10} \\\\ & \Rightarrow \frac{1}{n+1}\left[{ }^{n+1} C_1+{ }^{n+1} C_2+\ldots+{ }^{n+1} C_{n+1}\right]=\frac{1023}{10} \\\\ & \Rightarrow \frac{2^{n+1}-1}{n+1}=\frac{1023}{10}=\frac{2^{10}-1}{10} \\\\ & \Rightarrow n+1=10 \\\\ & \Rightarrow n=9 \end{aligned} $$