18) What is the sum of the numbers in the adjacent table?
(A) $\left(\frac{n(n+1)}{2}\right)^{2}$
(B) $n+\frac{n(n+1)^{2}}{2}$
(C) $n^{3}+n^{2}+n+1$
(D) $n(2 n-1)(n+1)$
(E) $n^{3}$.

| 1 | 2 | 3 | $\cdots$ | $n$ |
| :---: | :---: | :---: | :---: | :---: |
| 2 | 3 | 4 | $\cdots$ | $n+1$ |
| 3 | 4 | 5 | $\cdots$ | $n+2$ |
| $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ |
| $n$ | $n+1$ | $n+2$ | $\cdots$ | $2 n-1$ |
18) The answer is (E).

The numbers in the first row have a sum of $\frac{n(n+1)}{2}$, those in the second row $\frac{n(n+3)}{2}, \ldots$ and those in the last row

$$
\frac{n(2 n-1+n)}{2}=\frac{n(3 n-1)}{2} .
$$

Therefore, the sum of all the numbers is

$$
\frac{n}{2}((n+1)+(n+3)+(n+5)+\cdots+(3 n-1))=\frac{n}{2} \cdot n \cdot \frac{(n+1)+(3 n-1)}{2}=n^{3}
$$

## SECOND SOLUTION

Two numbers symmetric with respect to the diagonal formed by all $n$ have a sum of $2 n$. Therefore, the sum of the numbers in the table is the same as that obtained with a table consisting of all numbers equal to $n$, i.e., $n \cdot n^{2}=n^{3}$.