Question

If $\sum _{r=0}^n\left(\frac{r+2}{r+1}\right){}^n{C}_r=\frac{2^8-1}{6}$, then $n=$.

• A. $8$
• B. $4$
• C. $6$
• D. $5$

Answer 1

The correct option is D $5$

$\sum _{r=0}^n{\frac{r+2}{r+1}}^n{C}_r=\frac{2^8-1}{6}$

$\Rightarrow \sum _{r=0}^n[1+\frac{1}{r+1}]{}^n{C}_r=\frac{2^8-1}{6}$

$\Rightarrow 2^n+\sum _{r=0}^n\frac{1}{n+1}{.}^{n+1}{C}_{r+1}=\frac{2^8-1}{6}$

$\Rightarrow 2^n+\frac{2^{n+1}-1}{n+1}=\frac{2^8-1}{6}$

$\Rightarrow \frac{2^n(n+3)-1}{n+1}=\frac{2^8-1}{6}$

$\Rightarrow \frac{2^n(n+1+2)-1}{n+1}=\frac{2^5(6+2)-1}{6}$

On comparing, we get $n+1=6\Rightarrow n=5$