# 2002 AMC 10B Problems/Problem 23

## Problem 23

Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is $\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$

## Solution 1

First of all, write $a_3$ and $a_4$ in terms of $a_2.$ \begin{align*} a_3 &= a_{1+2} = 1 + a_2 + 2 = a_2 + 3\\ a_4 &= a_{2+2} = a_2 + a_2 + 4 = 2a_2 + 4 \end{align*} $a_6$ can be represented by $a_2$ in $2$ different ways. \begin{align*} a_{2+4} &= a_2 + a_4 + 8 = a_2 + 2a_2 + 4 + 8 = 3a_2 + 12\\ a_{3+3} &= 2a_3 + 9 = 6 + 2a_2 + 9 = 2a_2 + 15\\ \end{align*}

Since both are equal to $a_6,$ you can set them equal to each other. \begin{align*} 3a_2 + 12 &= 2a_2 + 15\\ a_2 &= 3 \end{align*}

Substitute the value of $a_2$ back into $a_6,$ and substitute that into $a_{12}.$ $$a_6 = 2a_2+15 = 6+15 = 21$$ $$a_{12} = a_{6+6} = a_6+a_6+36 = 21+21+36 = \boxed{\mathrm{(D) \ } 78}$$

## Solution 2

Substituting $n=1$ into $a_{m+n}=a_m+a_n+mn$: $a_{m+1}=a_m+a_{1}+m$. Since $a_1 = 1$, $a_{m+1}=a_m+m+1$. Therefore, $a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)$, and so on until $a_2 = a_1 + 2$. Adding the Left Hand Sides of all of these equations gives $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2$; adding the Right Hand Sides of these equations gives $(a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$. These two expressions must be equal; hence $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$ and $a_m = a_1 + (m + (m-1) + (m-2) + \cdots + 2)$. Substituting $a_1 = 1$: $a_m = 1 + (m + (m-1) + (m-2) + \cdots + 2) = 1+2+3+4+ \cdots +m = \frac{(m+1)(m)}{2}$. Thus we have a general formula for $a_m$ and substituting $m=12$: $a_{12} = \frac{(13)(12)}{2} = (13)(6) = \boxed{\mathrm{(D) \ } 78}$.

This is also the formula for the triangular numbers $T_n=1+2+3+...+n$, as seen in Solution 2
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