Starting from the right side, $$\begin{array}{rl} & \displaystyle\frac{1}{\binom{2n+1}{k}}+\frac{1}{\binom{2n+1}{k+1}} \\ \\ =& \displaystyle\frac{k!(2n+1-k)!}{(2n+1)!} +\frac{(k+1)!(2n+1-k-1)!}{(2n+1)!} \\ \\ =& \displaystyle\frac{k!(2n+1-k)! + (k+1)!(2n+1-k-1)! }{(2n+1)!} \\ \\ = & \displaystyle\frac{k!(2n-k)!((2n-k+1)+(k+1))}{(2n+1)!} \\ \\ =&\displaystyle\frac{k!(2n-k)!}{(2n)!}\cdot\frac{2n+2}{2n+1} \\ \\ =& \displaystyle\frac{2n+2}{2n+1}\cdot\frac{1}{\binom{2n}{k}} \end{array}$$