  Let $p$ be an odd prime and let
$f(x)=\sum_{i=1}^ka_ix^{p^{\alpha_i}+1}\in\Bbb F_{p^n}[x]$, where $0\le
\alpha_1<...<\alpha_k$. We consider the exponential sum $S(f,n)=\sum_{x\in\Bbb
F_{p^n}}e_n(f(x))$, where $e_n(y)=e^{2\pi i\text{Tr}_n(y)/p}$, $y\in\Bbb
F_{p^n}$, $\text{Tr}_n=\text{Tr}_{\Bbb F_{p^n}/\Bbb F_p}$. There is an
effective way to compute the nullity of the quadratic form
$\text{Tr}_{mn}(f(x))$ for all integer $m>0$. Assuming that all such nullities
are known, we find relative formulas for $S(f,mn)$ in terms of $S(f,n)$ when
$\nu_p(m) \le \min\{\nu_p(\alpha_i):1\le i\le k\}$, where $\nu_p$ is the
$p$-adic order. We also find an explicit formula for $S(f,n)$ when
$\nu_2(\alpha_1)=...= \nu_2(\alpha_k)<\nu_2(n)$. These results generalize those
by Carlitz and by Baumert and McEliece. Parallel results with $p=2$ were
obtained in a previous paper by the second author.
