1. Given $(a x+n)^{4}=a x^{4}+b x^{3}+c x^{2}+d x$
+16 . Then $b+c+d=$ $\qquad$
$=、 1.64$ or $\cdots 16$.
From the problem, we have $c^{\prime}=a, n^{\prime}=16$, then
$$
a=1, n= \pm 2 \text {. }
$$

When $a=2$,
$$
b+c+d=(1+2)^{4}-1-16=64 \text {; }
$$

When $n=-2$,
$$
b+c+d=(1-2)^{4}-1-16=-16 .
$$