2002 AMC 10B Problem 23

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 $\boxed{\textbf{(D)}\ 78}$.

A functional relation in two variables is usually far more information than needed; fixing one variable at the base case turns it into an ordinary one-step recursion. The increment $m + 1$ growing linearly is the signature of a quadratic closed form, and the triangular numbers are the first guess to test. The identity $T_{m+n} = T_m + T_n + mn$ is worth remembering.

Jump by doubling and one extra addition: $a_2 = a_1 + a_1 + 1 = 3$, $a_4 = 3 + 3 + 4 = 10$, $a_8 = 10 + 10 + 16 = 36$, then $a_{12} = a_8 + a_4 + 8\cdot 4 = 36 + 10 + 32 = 78$. Four steps, no formula.

• Using only m = n (doubling) and getting stuck because 12 is not a power of 2, or making an arithmetic slip while chaining twelve values by hand.
• Dropping the $mn$ term when specializing (writing $a_{m+1} = a_m + 1$), which gives $a_{12} = 12$ and matches no choice.