$1 \cdot 148$ Calculate $\sqrt{31 \times 30 \times 29 \times 28+1}$.
[Solution] Since
$$
\begin{aligned}
& (n-1) n(n+1)(n+2)+1 \\
= & \left(n^{2}+n\right)\left(n^{2}+n-2\right)+1 \\
= & \left(n^{2}+n\right)^{2}-2\left(n^{2}+n\right)+1 \\
= & \left(n^{2}+n-1\right)^{2},
\end{aligned}
$$

therefore, by setting $n=29$ in the above expression, we get
$$
28 \times 29 \times 30 \times 31+1=\left(29^{2}+29-1\right)^{2}=869^{2} \text {, }
$$

thus $\quad \sqrt{31 \times 30 \times 29 \times 28+1}=869$.