If (1)/(n+1)^nC_n+(1)/(n)^nC_n-1+ +(1)/(2)^nC_1+^nC_0=(1023)/(10) then n is equal to :

Question

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

Accepted Answer

$\frac{1}{n+1}{}^{n}{C}_{n}+\frac{1}{n}{}^{n}{C}_{n-1}+\ldots +\frac{1}{2}{}^{n}{C}_{1}+{}^{n}{C}_{0}=\frac{1023}{10}$

$\begin{gathered} \Rightarrow \sum _{r=0}^{n}\frac{1}{r+1}{}^{n}{C}_{r}=\frac{1023}{10} \ \, (\because {}^{n+1}{C}_{r+1}=\frac{n+1}{r+1}{}^{n}{C}_{r}) \ \Rightarrow \sum _{r=0}^{n}\frac{1}{n+1}{}^{n+1}{C}_{r+1}=\frac{1023}{10} \ \Rightarrow \frac{1}{n+1}[{}^{n+1}{C}_{1}+{}^{n+1}{C}_{2}+\ldots +{}^{n+1}{C}_{n+1}]=\frac{1023}{10} \ \Rightarrow \frac{{2}^{n+1}-1}{n+1}=\frac{1023}{10}=\frac{{2}^{10}-1}{10} \ \Rightarrow n+1=10 \ \Rightarrow n=9 \end{gathered}$

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