2002 AMC 10B Problem 23

The rule works for any $m$ and $n$, so choose values that make it simple. With $n = 1$ we can step from one term to the next.

1. Put $n = 1$ and use $a_1 = 1$:
$$a_{m+1} = a_m + a_1 + m \cdot 1 = a_m + (m + 1).$$
In words: to get the next term, add the next counting number.
2. Build the terms:

3. The pattern: $a_m = 1 + 2 + \cdots + m$. So
$$a_{12} = 1 + 2 + \cdots + 12 = \frac{12 \cdot 13}{2} = 78.$$

The answer is $\boxed{\textbf{(D)}\ 78}$.

A rule that holds "for all $m$ and $n$" gives far more than you need. Fixing one of the letters at the smallest value turns it into a step-by-step rule, and then you just build the sequence. Recognizing $1, 3, 6, 10, \ldots$ as sums of counting numbers lets you jump straight to the $12$th term with the formula $\frac{m(m+1)}{2}$.

Take bigger jumps. Using $m = n$ doubles the position:
$$a_2 = 1 + 1 + 1 = 3, \qquad a_4 = 3 + 3 + 2 \cdot 2 = 10, \qquad a_8 = 10 + 10 + 4 \cdot 4 = 36.$$
Then use $m = 8$, $n = 4$: $a_{12} = 36 + 10 + 8 \cdot 4 = 78$.

The relation holds for every pair $(m, n)$, so specialize to $n = 1$:
$$a_{m+1} = a_m + a_1 + m\cdot 1 = a_m + (m + 1).$$
So going from index $m$ to index $m + 1$ adds $m + 1$. Starting from $a_1 = 1$:
$$a_2 = 1 + 2 = 3,\quad a_3 = 3 + 3 = 6,\quad a_4 = 6 + 4 = 10,\quad \ldots$$
In general $a_m = 1 + 2 + \cdots + m = \frac{m(m+1)}{2}$, the triangular numbers. (Check: $\frac{(m+n)(m+n+1)}{2} - \frac{m(m+1)}{2} - \frac{n(n+1)}{2} = mn$, so the formula really satisfies the relation.)

Therefore
$$a_{12} = \frac{12 \cdot 13}{2} = 78.$$

The answer is $\boxed{\textbf{(D)}\ 78}$.