Question

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

• A. 8
• B. 4
• C. 6
• D. 5

Answer 1

Answer: (D)

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