Suma de la ponderación de los cuadrados de los coeficientes binomiales

Question

Mostrar que

$$\sum_{k=0}^n \left[ \frac{n-2k}{n} {n\choose k}\right]^2=\frac{2}{n}{2n-2 \choose n-1}.$$

Answer 1

$$ \begin{align} \sum_{k=0}^n\left[\frac{n-2k}{n}\binom{n}{k}\right]^2 &=\sum_{k=0}^n\left[\binom{n}{k}-2\binom{n-1}{k-1}\right]^2\\ &=\sum_{k=0}^n\left[\binom{n-1}{k}-\binom{n-1}{k-1}\right]^2\\ &=\sum_{k=0}^n\binom{n-1}{k}^2+\binom{n-1}{k-1}^2-2\binom{n-1}{k}\binom{n-1}{k-1}\\ &=2\binom{2n-2}{n-1}-2\sum_{k=0}^n\binom{n-1}{k}\binom{n-1}{n-k}\\ &=2\binom{2n-2}{n-1}-2\binom{2n-2}{n}\\ &=2\binom{2n-2}{n-1}-2\frac{n-1}{n}\binom{2n-2}{n-1}\\ &=\frac2n\binom{2n-2}{n-1} \end{align} $$