Mathematics - Binomial Theorem Question with Solution | TestHub

Given that $\frac{1}{n+1}{ }^n{C}_n+\frac{1}{n}{ }^n{C}_{n-1}+...+\frac{1}{2}{ }^n{C}_1{+}^n{C}_0=\frac{1023}{10}$

$\Rightarrow \sum _{r=0}^n\frac{1}{r+1}{ }^n{C}_r=\frac{1023}{10} \left(\therefore { }^{n+1}{C}_{r+1}=\frac{n+1}{r+1}{ }^n{C}_r\right)$

$\Rightarrow \sum _{r=0}^n\frac{1}{n+1}{ }^{n+1}{C}_{r+1}=\frac{1023}{10}$

$\Rightarrow \frac{1}{n+1}\left[{}^{n+1}{C}_1{+}^{n+1}{C}_2+...{+}^{n+1}{C}_{n+1}\right]=\frac{1023}{10}$

We know that ${}^{n}C_0+{}^{n}C_1+{}^{n}C_2+.......+{}^{n}C_n=2^n$

$\Rightarrow \frac{2^{n+1}-1}{n+1}=\frac{1023}{10}=\frac{2^{10}-1}{10}$

$\Rightarrow n+1=10$

$\Rightarrow n=9$

Hence this is the correct option.