The Problems of Ukrainian Authors

Since $\frac{n\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)}{k+1}=\frac{n(n-1)!}{(k+1)k!(n-k-1)!}=\frac{n!}{k!(n-k-1)!}=\left(\genfrac{}{}{0ex}{}{n}{k+1}\right)$, the left-hand-side sum takes the form $$\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\left(\genfrac{}{}{0ex}{}{n}{k+1}\right)=\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\left(\genfrac{}{}{0ex}{}{n}{n-k-1}\right)=\left(\genfrac{}{}{0ex}{}{2n-1}{n}\right).$$ The latter equality follows from the properties of binomial coefficients. On the other hand, $$\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=\left(\genfrac{}{}{0ex}{}{2n-1}{n-1}\right)+\left(\genfrac{}{}{0ex}{}{2n-1}{n}\right)=2\left(\genfrac{}{}{0ex}{}{2n-1}{n-1}\right),$$ which finishes the proof.