How can I show that $\sum_{k=0}^n [\frac{(n-2k)}{n}{n \choose k}]^2 = \frac{2}{n}{2n - 2 \choose n - 1}$? I want to show this
$\sum_{k=0}^n [\frac{(n-2k)}{n}{n \choose k}]^2 = \frac{2}{n}{2n - 2 \choose n - 1}$
I tried to solve it using this identities:
I) $\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}$
II) ${n \choose k} = {n-1 \choose k-1}\frac{n}{k}$
III) ${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$
Any hints?
Edit: I could solve it with a hint from TMM and the fact that $\sum_{k=0}^n {s \choose k}{r \choose n-k} = {s+r \choose n}$
 A: Solution
$\sum_{k=0}^{n}[\frac{n-2k}{n}\binom{n}{k}]^{2}$ 
$= \frac{1}{n^{2}}\sum_{k=0}^{n}(n-2k)^2\binom{n}{k}^2 $ 
$=\frac{1}{n^{2}}[\sum_{k=0}^{n}n^2\binom{n}{k}^2 + 4\sum_{k=0}^{n}k^2\binom{n}{k}^2 - 4\sum_{k=0}^{n}nk\binom{n}{k}^2] $
$= \sum_{k=0}^{n}\binom{n}{k}^2 + \frac{4}{n^{2}}\sum_{k=0}^{n}n^2\binom{n-1}{k-1}^2 - \frac{4}{n^{2}}\sum_{k=0}^{n}n^2\binom{n-1}{k-1}\binom{n}{k}$ , here we use the fact that ${n \choose k} = {n-1 \choose k-1}\frac{n}{k}$ so $k{n \choose k} = n{n-1 \choose k-1}$
$= \sum_{k=0}^{n}\binom{n}{k}^2 + 4\sum_{k=0}^{n}\binom{n-1}{k-1}^2 - 4\sum_{k=0}^{n}\binom{n-1}{n-k}\binom{n}{k}$ , we use in the last sum: $\binom{n}{k}=\binom{n}{n-k}$ so $\binom{n-1}{k-1}=\binom{n-1}{n-k}$
Finally, knowing that:
$\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}$ and $\sum_{k=0}^n {s \choose k}{r \choose n-k} = {s+r \choose n}$
$= \binom{2n}{n} + 4\binom{2n-2}{n-1} - 4\binom{2n-1}{n} = \binom{2n}{n} + 4(\binom{2n-2}{n-1} - \binom{2n-1}{n})$
$= \binom{2n}{n} - 4\binom{2n-2}{n}$ , because ${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$ then $ -{n-1 \choose k-1}={n-1 \choose k}-{n \choose k}$ in this case $\binom{2n-2}{n-1} - \binom{2n-1}{n} = -\binom{2n-2}{n}$
$= \frac{(2n)!}{(n!)^2}-2^2\frac{(2n-2)!}{n!(n-2)!}$
$= \frac{(2n)!-2^2(2n-2)!n(n-1)}{(n!)^2}$
$= \frac{(2n)!-(2n-2)!(2n)2(n-1)}{(n!)^2} = \frac{(2n)!-(2n-2)!(2n)(2n-1 - 1)}{(n!)^2}$
$= \frac{(2n)!-(2n-2)!(2n)(2n-1) -(2n-2)!(2n)(-1) }{(n!)^2}$
$= \frac{(2n)!-(2n)! +(2n-2)!(2n) }{(n!)^2}$
$= \frac{(2n-2)!}{((n-1)!)^2}\frac{2}{n}$
$= \frac{2}{n}\binom{2n-2}{n-1}$
