Question

If $\sum _{r=0}^n\left(\frac{r+2}{r+1}\right)C_r=\frac{2^8-1}{6}$, where $C_r={}^n{C}_r$, then the value of $n$ is

• A. $8$
• B. $6$
• C. $5$
• D. $4$

Answer 1

The correct option is C $5$

$\sum _{r=0}^n\left(\frac{r+1+1}{r+1}\right)C_r=\sum _{r=0}^n{C}_r+\sum _{r=0}^n\frac{(n+1)}{(r+1)(n+1)}{C}_r=\sum _{r=0}^n{}^n{C}_r+\frac{1}{n+1}\sum _{r=0}^n{}^{n+1}{C}_{r+1}=2^n+\frac{1}{n+1}(2^{n+1}-1)\Rightarrow \sum _{r=0}^n\left(\frac{r+2}{r+1}\right)C_r=\frac{2^8-1}{6}\Rightarrow 2^n+\frac{1}{n+1}(2^{n+1}-1)=\frac{2^8-1}{6}\Rightarrow 2^n+\frac{2^{n+1}}{n+1}-\frac{1}{n+1}=\frac{2^8}{6}-\frac{1}{6}$
By observation, $n=5$
Checking for $n=5$, we get
$2^5+\frac{2^6}{6}-\frac{1}{6}=\frac{2^5(6+2)}{6}-\frac{1}{6}=\frac{2^8-1}{6}$

Hence, $n=5$