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

Count the total number of terms through the end of the

k

th block

Find the last complete block before term

1234

, then account for the partial next block

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.

Problem 24 in other years