Question

设 $\{a_k\}$ 是整数序列，满足 $a_1 = 1$ ，且对所有正整数 $m$ 、 $n$ 都有 $a_{m+n} = a_m + a_n + mn$ 。求 $a_{12}$ 。

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 $n.$ What is $a_{12}?$

$45$

$56$

$67$

$78$

$89$

令 $n = 1$ ，得到 $a_{m+1} = a_m + a_1 + m$ $= a_m + (m + 1)$ ，所以 $a_{m+1} - a_m = m + 1$ 。

从 $m = 1$ 加到 $11$ ，有 $\begin{aligned} a_{12} - a_1 &= 2 + 3 + \cdots + 12 \\ &= \dfrac{12\cdot 13}{2} - 1 \\ &= 77. \end{aligned}$

$a_{12} = 1 + 77 = 78$ 。

所以正确答案是 D。

Setting $n = 1,$ we get $a_{m+1} = a_m + a_1 + m$ $= a_m + (m + 1),$ so $a_{m+1} - a_m = m + 1.$

Summing from $m = 1$ to $11,$ $\begin{aligned} a_{12} - a_1 &= 2 + 3 + \cdots + 12 \\ &= \dfrac{12\cdot 13}{2} - 1 \\ &= 77. \end{aligned}$

Therefore $a_{12} = 1 + 77 = 78.$

Thus, the correct answer is D.