<s>[INST] Show that $\sum_{k=0}^{n+1}\left(\binom{n}{k}-\binom{n}{k-1}\right)^2 = \frac{2}{n+1}\binom{2n}{n}$ 
Show that $$\displaystyle\sum_{k=0}^{n+1}\left(\dbinom{n}{k}-\dbinom{n}{k-1}\right)^2 = \dfrac{2}{n+1}\dbinom{2n}{n}$$ where  $n \in \mathbb{N}$

Consider $\dbinom{n}{r}=0 $ for $r<0 $ and $r>n$. Using $\displaystyle\sum_{k=0}^{n}\dbinom{n}{k}^2 = \dbinom{2n}{n}$, we get $\displaystyle\sum_{k=0}^{n+1}\left(\dbinom{n}{k}-\dbinom{n}{k-1}\right)^2 = \displaystyle\sum_{k=0}^{n+1} \dbinom{n}{k}^2 + \displaystyle\sum_{k=0}^{n+1} \dbinom{n}{k}^2 - 2\displaystyle\sum_{k=0}^{n+1}\dbinom{n}{k}\dbinom{n}{k-1} \\ \hspace{47mm} = 2\dbinom{2n}{n}-2\displaystyle\sum_{k=1}^{n}\dbinom{n}{k}\dbinom{n}{k-1} \\ \hspace{47mm} = 2\dbinom{2n}{n}-2\displaystyle\sum_{k=1}^{n}\dfrac{1}{k(n-k+1)}\dbinom{n}{k-1}^2 \\ \hspace{47mm} = 2\dbinom{2n}{n}-\dfrac{2}{n+1}\displaystyle\sum_{k=1}^{n}\left(\dfrac{1}{k}+\dfrac{1}{n-k+1} \right)\dbinom{n}{k-1}^2 $ 
Now I am stuck at this point, how do I further progress in this proof? Is there a better way to solve this problem ?
 [/INST] The given sum is equal to the constant term of $f(z)f(1/z)$, where $$f(z)=\sum_{k=0}^{n+1}\left[\binom{n}{k}-\binom{n}{k-1}\right]z^k=(1-z)(1+z)^n;$$ in other words, it is the coefficient of $z^{n+1}$ in $$z^{n+1}f(z)f(1/z)=-(1-z)^2(1+z)^{2n},$$ equal to $-\binom{2n}{n+1}+2\binom{2n}{n}-\binom{2n}{n-1}$ which reduces to the expected result.
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