72 For any region $D$ on the Cartesian plane, let $N(D)$ denote the number of all lattice points in $D$, and $n$ is a natural number greater than 1.
(1) Let $A$ be the region enclosed by the curve $y=x^{2}$ and the lines $x=n$ and $y=1$ (including the boundaries). Find $N(A)$.
(2) Let $B$ be the region in the first quadrant enclosed by the curve $y=x^{2}$ and the lines $x=1$ and $y=n^{2}$ (including the boundaries). Find $N(B)$.
(3) Find $\sum_{k=1}^{n^{2}}[\sqrt{k}]$.
72 (1) On the line $x=k$, there are $k^{2}$ lattice points $(k, 1),(k, 2), \cdots,\left(k, k^{2}\right)$, so
$$
N(A)=1^{2}+2^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1)
$$
(2) Clearly, $N(A \cup B)=n^{3}$. The lattice points in $A \cap B$ are $\left(k, k^{2}\right)(k=1,2, \cdots, n)$, so $N(A \cap B)=n$. Since
$$
N(A \cup B)=N(A)+N(B)-N(A \cap B),
$$

thus
$$
N(B)=N(A \cup B)-N(A)+N(A \cap B)=\frac{1}{6}\left(4 n^{3}-3 n^{2}+5 n\right) .
$$
(3) The intersection points of $y=k\left(k=1,2, \cdots, n^{2}\right)$ and the curve $y=x^{2}(x>0)$ have the abscissa $x=\sqrt{k}$, and the integer part $[\sqrt{k}]$ is the number of lattice points in region $B$ on the line $y=k$. Therefore,
$$
\sum_{k=1}^{n^{2}}[\sqrt{k}]=N(B)=\frac{1}{6}\left(4 n^{3}-3 n^{2}+5 n\right) .
$$