Firstly, it can be verified that  $$\frac{k+1}{\binom{2n}{k}}=\frac{2n+1}{\binom{2n+1}{k+1}}=\frac{2n+1}{\binom{2n+1}{2n-k}}=\frac{2n-k}{\binom{2n}{2n-k-1}}$$

Therefore, $$\begin{align*} k=0 &\implies & \frac{1}{\binom{2n}{0}} &= \frac{2n}{\binom{2n}{2n-1}}\\ \\ k=1 &\implies &\frac{2}{\binom{2n}{1}} &= \frac{2n-1}{\binom{2n}{2n-2}}\\ \cdots\\ k=n-1 &\implies &\frac{n}{\binom{2n}{n-1}} &=\frac{n+1}{\binom{2n}{n}}\end{align*}$$

Adding these relations will lead to the desired result.