Question

If $\sum\limits^{n}_{r=0} \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)

$\displaystyle\sum_{r=0}^{n} \frac{r+2}{r+1}^{n} C_{r}=\frac{2^{8}-1}{6}$
$\Rightarrow \displaystyle\sum_{r=0}^{n}\left[1+\frac{1}{r+1}\right]^{n} C_{r}=\frac{2^{8}-1}{6}$
$\Rightarrow 2^{n}+\displaystyle\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 $
$\Rightarrow n =5$