1996 AMC 12 Problem 24

Attempt Problem 24 of the 1996 AMC 12 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 1996 AMC 12 solutions, or check the answer key.

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24.

The sequence $\begin{gathered} 1,2,1,2,2,1,2,2,2,1,2,\\ 2,2,2,1,2,2,2,2,2,1,2,\ldots \end{gathered}$ consists of $1$ ’s separated by blocks of $2$ ’s with $n$ $2$ ’s in the $n$ th block. The sum of the first $1234$ terms of this sequence is

$1996$

$2419$

$2429$

$2439$

$2449$

Answer: B

Concepts:sequences triangular numbers

Difficulty rating: 1860

Small Hint:

Count the total number of terms through the end of the $k$ th block

Big Hint:

Find the last complete block before term $1234$ , then account for the partial next block

Solution:

Through block $k$ there are $k$ ones and $\frac{k(k+1)}{2}$ twos, hence $\frac{k(k+3)}{2}$ terms. For $k=48$ this is $1224.$ Their sum is $48+2\left(\frac{48\cdot49}{2}\right)=2400.$ The next $10$ terms are one $1$ and nine $2$ ’s, with sum $19.$ The requested sum is $2419,$ so the correct answer is B.

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