52 Given $\left(a x^{4}+b x^{3}+c x^{2}+d x+e\right)^{5} \cdot\left(a x^{4}-b x^{3}+c x^{2}-d x+e\right)^{5}=a_{0}+a_{1} x+$ $a_{2} x^{2}+\cdots+a_{41} x^{10}$, then $a_{1}+a_{3}+a_{5}+\cdots+a_{39}=$ $\qquad$.
520 . Let $F(x)=\left(a x^{4}+b x^{3}+c x^{2}+d x+e\right)^{5} \cdot\left(a x^{4}-b x^{3}+c x^{2}-d x+e\right)^{5}$, from $F(-x)=F(x)$ we know that $F(x)$ is an even function, so the expansion of $F(x)$ does not contain odd powers of $x$, i.e.,
$$
a_{1}=a_{3}=\cdots=a_{39}=0,
$$

thus
$$
a_{1}+a_{3}+a_{5}+\cdots+a_{39}=0 .
$$