Mathematics - Binomial Theorem Question with Solution | TestHub

${\sum }_{r=0}^n\frac{r+2}{r+1}{}^n{C}_r=\frac{2^8-1}{6}$ $$\begin{array}{rl}& \Rightarrow \sum _{r=0}^n\left[1+\frac{1}{r+1}\right]{}^n{C}_r=\frac{2^8-1}{6}\\ & \Rightarrow 2^n+\sum _{r=0}^n\frac{1}{n+1}\cdot {}^{n+1}{C}_{r+1}=\frac{2^8-1}{6}\\ \Rightarrow 2^n+& \frac{2^{n+1}-1}{n+1}=\frac{2^8-1}{6}\Rightarrow \frac{2^n(n+3)-1}{n+1}=\frac{2^8-1}{6}\\ \Rightarrow & \frac{2^n(n+1+2)-1}{n+1}=\frac{2^5(6+2)-1}{6}\end{array}$$ Comparing we get $\mathrm{n}+1=6\Rightarrow \mathrm{n}=5$