Ta có: $(x+1)^n=\sum_{i=0}^nC_n^ix^i \Rightarrow \sum_{i=0}^nC_n^i=2^n$
Lại có:
$\dfrac{C_n^i}{i+1}=\dfrac{n!}{i!(n-i)!(i+1)}=\dfrac{(n+1)!}{(i+1)!(n-i)!(n+1)}=\dfrac{1}{n+1}C_{n+1}^{i+1}$
Suy ra: $\sum_{i=1}^n\dfrac{C_n^i}{i+1}=\dfrac{1}{n+1}\sum_{i=1}^nC_{n+1}^{i+1}=\dfrac{1}{n+1}(2^{n+1}-C_{n+1}^0-C_{n+1}^1)=\dfrac{1}{n+1}(2^{n+1}-n-2)$