AMC10 2002 B

AMC10 2002 B · Q23

AMC10 2002 B · Q23. It mainly tests Sequences & recursion (algebra), Patterns & sequences (misc).

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

设 $\{a_k\}$ 为整数序列，满足 $a_1 = 1$ 且 $a_{m+n} = a_m + a_n + mn$，对所有正整数 $m$ 和 $n$ 成立。则 $a_{12}$ 是

(A) 45 45

(B) 56 56

(D) 78 78

(E) 89 89

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) By setting $n=1$ in the given recursive equation, we obtain $a_{m+1}=a_m+a_1+m$, for all positive integers $m$. So $a_{m+1}-a_m=m+1$ for each $m=1,2,3,\ldots$ Hence, $a_{12}-a_{11}=12,\ a_{11}-a_{10}=11,\ \ldots,\ a_2-a_1=2.$ Summing these equalities yields $a_{12}-a_1=12+11+\cdots+2$. So $a_{12}=12+11+\cdots+2+1=\dfrac{12(12+1)}{2}=78.$ OR We have $a_2=a_{1+1}=a_1+a_1+1\cdot1=1+1+1=3,$ $a_3=a_{2+1}=a_2+a_1+2\cdot1=3+1+2=6,$ $a_6=a_{3+3}=a_3+a_3+3\cdot3=6+6+9=21,$ and $a_{12}=a_{6+6}=a_6+a_6+6\cdot6=21+21+36=78.$

（D）在给定的递推式中令 $n=1$，得到对所有正整数 $m$，有 $a_{m+1}=a_m+a_1+m$。因此对每个 $m=1,2,3,\ldots$，有 $a_{m+1}-a_m=m+1$。于是， $a_{12}-a_{11}=12,\ a_{11}-a_{10}=11,\ \ldots,\ a_2-a_1=2。$ 把这些等式相加得 $a_{12}-a_1=12+11+\cdots+2$。所以 $a_{12}=12+11+\cdots+2+1=\dfrac{12(12+1)}{2}=78。$ 或者我们有 $a_2=a_{1+1}=a_1+a_1+1\cdot1=1+1+1=3，$ $a_3=a_{2+1}=a_2+a_1+2\cdot1=3+1+2=6，$ $a_6=a_{3+3}=a_3+a_3+3\cdot3=6+6+9=21，$ 并且 $a_{12}=a_{6+6}=a_6+a_6+6\cdot6=21+21+36=78。$

Topics