I have this identity that I'd like to prove.
$$\displaystyle{\sum_{k=0}^{n}\bigg(\dfrac{n-2k}{n}\binom{n}{k}}\bigg)^2=\dfrac{2}{n}\binom{2n-2}{n-1}$$
Here's what I have done so far: (using a binomial indentity)
$$=\displaystyle{\sum_{k=0}^{n}\bigg({\binom{n}{k}-2\binom{n-1}{k-1}}\bigg)^2}$$
$$=\displaystyle{\sum_{k=0}^{n}\bigg({\binom{n-1}{k}-\binom{n-1}{k-1}}\bigg)^2}$$
$$=\displaystyle{\sum_{k=0}^{n}\bigg({\binom{n-1}{k}-\binom{n-1}{k-1}}\bigg)^2}$$
At this point I expanded the square, Here's where I made a mistake
$$=\displaystyle{{\sum_{k=0}^{n}\binom{n-1}{k}^2+\sum_{k=0}^{n}\binom{n-1}{k-1}}^2-\sum_{i=0}^{n}\sum_{j=0}^{n}}\binom{n-1}{j-1}\binom{n-1}{i}$$
$$=\displaystyle{{\sum_{k=0}^{n}\binom{n-1}{k}^2+\sum_{k=0}^{n}\binom{n-1}{k-1}}^2-\sum_{i=0}^{n}\sum_{j=0}^{n}}\binom{n-1}{j-1}\binom{n-1}{i}$$
$$=\displaystyle{2\binom{2n-2}{n-1}-\sum_{i=0}^{n}\sum_{j=0}^{n}}\binom{n-1}{j-1}\binom{n-1}{i}$$
because I can separate the sums
$$=\displaystyle{2\binom{2n-2}{n-1}-2^{2n-2}}$$
Clearly, at some point here I made a stupid mistake. I was hoping someone will point the error to me and perhaps give me a hint. I prefer hints to complete solutions. Thank you for your time.
Best Answer
After expanding the square, the summation index should be the same for the last summation, making it a single summation over $i,$ not a double summation with $i,j.$ You then have a sum which can be evaluated with Vandermonde's identity.