2021/22 AHSMC Part 1 Problem 7

Problem 7 of 16

In the sequence $1, 4, 1, 4, 4, 1, 4, 4, 4, 1, 4, 4, 4, 4, 1, \ldots$ there are $1$ 's separated by strings of $4$ 's, where each string of $4$ 's is one longer than the previous string. The value of $n$ for which the first $n$ terms sum up to $235$ is

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Solution

If we add up $1+4+1+(4+4)+1+(4+4+4)+\cdots+1+(4+\cdots+4)$ where there are $k$ $4$ 's in the last string, we get $k + 4k(k+1)/2 = 2k^2 + 3k$ . When $k = 10$ this equals $230$ which is close to $235$ , so to get exactly $235$ we need to add in another $1$ and one $4$ , and the number of terms in the sequence is then $10 + (1+\cdots+10) + 2 = 10 + 55 + 2 = 67$ . The answer is (C).

University of Alberta

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