# Thread: Binomial coefficients 4

1. ## Binomial coefficients 4

*Show that

$
\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}^2=(-1)^n\binom{2n}{n}
$

2. Hi,

$\sum_{k=0}^r (-1)^k\binom{r}{k}x^{2k} = (1-x^2)^r = (1-x)^r(1+x)^r = \sum_{j=0}^r (-1)^j\binom{r}{j}x^{j}\cdot \sum_{k=0}^r \binom{r}{k}x^{k}=$
$=\sum_{k=0}^{2r}\left( \sum_{j=0}^k (-1)^j \binom{r}{j}\binom{r}{k-j} \right)x^k$

Looking at coefficients at $x^{2n}$ we get

$\sum_{j=0}^{2n} (-1)^j \binom{r}{j}\binom{r}{2n-j}= (-1)^n\binom{r}{n}$

Special case of this is when you take $r=2n$:

$\sum_{j=0}^{2n} (-1)^j \binom{2n}{j}\binom{2n}{2n-j}= (-1)^n\binom{2n}{n}$

and using the identity $\binom{n}{k}=\binom{n}{n-k}$ gives you the desired result.