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

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2002 AMC 10B Problem 23

Below is the professionally curated solution for Problem 23 of the 2002 AMC 10B, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:functional equation telescoping triangular number

Difficulty rating: 1510

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 $a_{12}?$

$45$

$56$

$67$

$78$

$89$

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 $11,$ $a_{12} - a_1 = 2 + 3 + \cdots + 12 = \dfrac{12\cdot 13}{2} - 1 = 77.$

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

Thus, the correct answer is D.

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