2002 AMC 10B Problem 23 — Solution | LIVE by Po-Shen Loh

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2002 AMC 10B Problem 23

Attempt Problem 23 of the 2002 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B solutions, or check the answer key.

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

$45$

$56$

$67$

$78$

$89$

Answer: D

Concepts:functional equation telescoping triangular number

Difficulty rating: 1510

Solution:

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.

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