Ta có: $(2t+1)^n=\sum_{k=0}^{n}C^k_n(2t)^k$
$\Rightarrow (2t+1)^n=\sum_{k=0}^{n}2^kC^k_nt^k$
$\Rightarrow \int\limits_0^1(2t+1)^ndt=\int\limits_0^1\sum_{k=0}^{n}2^kC^k_nt^kdt$
$\Rightarrow \dfrac{(2t+1)^{n+1}}{2(n+1)}\left|\begin{array}{l}1\\0\end{array}\right.=\sum_{k=0}^n\dfrac{2^kC^k_nt^{k+1}}{k+1}\left|\begin{array}{l}1\\0\end{array}\right.$
$\Rightarrow \sum_{k=0}^n\dfrac{2^kC^k_n}{k+1}=\dfrac{3^{n+1}-1}{2(n+1)}$