Question

$\displaystyle\sum_{k=0}^n \frac{^nC_k}{k+1} = $

• A. $ \frac{2^n -1}{n+1}$
• B. $ \frac{2^{n-1} -1}{n - 1}$
• C. $ \frac{2^{n + 1} -1}{n + 1}$
• D. $ \frac{2^n -1}{n}$

Answer 1

Answer: (C)

$\displaystyle\sum_{k=0}^{n} \frac{^{n}C_{k}}{k+1} = \displaystyle\sum^{n}_{k=0} \frac{n!}{\left(k+1\right)k! \left(n -k\right)!}$
$= \displaystyle\sum_{k=0}^{n} \frac{n!\left(n+1\right)}{\left(k+1\right)!\left(n-k\right)!\left(n+1\right)} $
$= \frac{1}{n+1} \displaystyle\sum _{k=0}^{n} \frac{\left(n+1\right)!}{\left(k+1\right)!\left(n+1-k-1\right)!} $
$ = \frac{1}{n+1} \displaystyle\sum_{k = 0}^{n} \,{}^{n+1}C_{k +1} = \frac{1}{n+1} \left[ 2^{n+1} - 1\right]$