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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

Small hint:

Set

n = 1

to get

a_{m+1} = a_m + a_1 + m

Big hint:

So

a_{m+1} - a_m = m + 1

; sum these differences up to

a_{12}

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.