Alternate binomial sum

Prove that:

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

Solution



Using the binomial expansion for $(1-x^2)^{2n}$ we have that:

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

On the other hand we have that:

$$\begin{aligned}
\left ( 1-x^2 \right )^{2n} &=\left ( 1-x \right )^{2n}\left ( 1+x \right )^{2n} \\
 &= \left [ \sum_{i=0}^{2n}\binom{2n}{i}(-1)^i x^i \right ]\left [ \sum_{j=0}^{2n}\binom{2n}{j}x^j \right ]\\
 &= \sum_{i, j} \binom{2n}{i}\binom{2n}{j}(-1)^i x^{i+j}\\
 &=\sum_{k=0}^{4n}x^k \sum_{i+j=k} (-1)^i \binom{2n}{i}\binom{2n}{j}
\end{aligned}$$

Equating the coefficients of $x^{2n}$ we get:

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

as wanted.