Solve the following summation $S = \dfrac{n \choose 0}{1} + \dfrac{n \choose 1}{2} +  
 \dfrac{n \choose 2}{3}+\dotsb+\dfrac{n \choose n}{n+1}$

$\bf{My\; Solution::}$ We can write $$\displaystyle S = \sum_{k=0}^{n}\binom{n}{k}\cdot \frac{1}{k+1} = \sum_{k=0}^{n}\frac{n!}{k!\cdot (n-k)!}\cdot \frac{1}{k+1}$$
$$\displaystyle S = \frac{1}{n+1}\sum_{k=0}^{n}\frac{(n+1)!}{(k+1)!\cdot (n-k)!}=\frac{1}{n+1}\sum_{k=0}^{n}\binom{n+1}{k+1} = \frac{2^{n+1}-1}{n+1}.$$
Above we have used the Identity $$\displaystyle \sum_{k=0}^{n}\binom{n}{k} = 2^n$$.
So for $$\displaystyle \sum_{k=0}^{n}\binom{n+1}{k+1} = 2^{n+1}$$
