<s>[INST] The summation $\sum_{k=0}^n \frac{{2n \choose k}}{k+1}$ Mathematica gives this summation
$$\sum_{k=0}^n \frac{{2n \choose k}}{k+1}$$
in terms of a regularized Gauss hypergeometric function $~_2F_1$. Can the hypergeometric function be eliminated to have a simple answer?
 [/INST] If you're looking for a (somewhat) closed form solution, observe that
\begin{aligned}
\frac{\binom{2n}{k}}{k+1} &= \frac{(2n)!}{(k+1)! \cdot (2n - k)!} \\
&= \frac{1}{2n+1} \frac{(2n+1)!}{(k+1)! \cdot ((2n+1) - (k+1))!} \\
&= \frac{1}{2n+1}\binom{2n+1}{k+1}.
\end{aligned}
So, by the Binomial theorem, we have
\begin{aligned}
\sum_{k=0}^n \frac{\binom{2n}{k}}{k+1} &=  \frac{1}{2n+1} \sum_{k=0}^n  \binom{2n+1}{k+1} \\
&=  \frac{1}{2n+1} \left(\sum_{k=0}^{n+1}  \binom{2n+1}{k} - 1 \right) \\
&=  \frac{1}{2n+1} \left(\frac{2^{2n+1}}{2} + \binom{2n+1}{n+1} - 1 \right) \\
&=  \frac{1}{2n+1} \left(2^{2n} + \binom{2n+1}{n}- 1 \right).
\end{aligned}
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