29 Let $f(n)=\left(C_{n}^{2}\right)^{2} \cdots\left(C_{n}^{1}\right)^{2}+\left(C_{n}^{2}\right)^{2}-\left(C_{n}^{3}\right)^{2}+\cdots+(-1)^{k}\left(C_{n}^{k}\right)^{2}+\cdots+$ $(-1)^{n}\left(C_{n}^{n}\right)^{2}$ (where $n$ is even), then $f(1990)=$ $\qquad$ .
$29-\mathrm{C}_{1999}^{695}$
$$
(1+x)^{n}=\sum C_{, \ldots}^{*} x^{n+k} \cdot(1-x)^{n}:=\sum C_{m}^{*}(-1)^{k} x^{k}
$$

So $\sum(-1)^{k}\left(C_{n}^{k}\right)^{2}$ represents the coefficient of $x^{n}$ in $(1+x)^{n}(1-x)^{n}$. The coefficient of $x^{n}$ in $\left(1-x^{2}\right)^{n}$ is $C_{n}^{\frac{n}{2}} \cdot(-1)^{\frac{n}{2}}$, so
$$
f_{(1990)}=-C_{1990}^{955}
$$