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 ?
 A: The way you simplify $\displaystyle{\sum_{k=0}^{n+1}\binom nk^2}$ also works for $\displaystyle{\sum_{k=0}^{n+1}\binom nk \binom n{k-1}}$.
Rewriting $\binom n{k-1} = \binom n{n-k+1}$ we can turn the second sum into $$\sum_{i+j=n+1} \binom ni \binom nj.$$ This is the coefficient of $x^{n+1}$ in $(1+x)^n (1+x)^n$, or $\binom{2n}{n+1}$ - but the generating function approach has already been done in the other answer to this question. We can also think of this sum as counting the number of lattice paths from $(0,0)$ to $(n+1,n-1)$ by partitioning them according to which point $(i,j-1)$ with $i+j=n+1$ they pass through: there are $\binom ni$ ways to get from $(0,0)$ to $(i,j-1)$ and $\binom n{n-j}=\binom nj$ ways to get from $(i,j-1)$ to $(n+1,n-1)$.
Anyway, once you arrive at $$\sum_{k=0}^{n+1} \binom{n}{k}^2 + \sum_{k=0}^{n+1} \binom{n}{k}^2 - 2\sum_{k=0}^{n+1}\binom{n}{k}\binom{n}{k-1}$$ we can handle all three sums in this way: you've already done the first two and my previous paragraph handles the third. We get $$\binom{2n}{n} + \binom{2n}{n} - 2\binom{2n}{n+1}$$ and since $\binom{2n}{n+1} = \frac{n}{n+1} \binom{2n}{n}$, this simplifies to $(1 + 1 - \frac{2n}{n+1}) \binom{2n}{n} = \frac{2}{n+1} \binom{2n}{n}$, as desired.
By the way, the general identity at work here is Vandermonde's identity $$\sum_{i+j=k} \binom ni \binom mj = \binom{n+m}{k}.$$
A: 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.
