Competition problem — Number theory

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Number theory Difficulty 4.9 AIME Find the answer

9. The sequence $1,1,2,1,1,3,1,1,1,4,1,1,1,1$ , $5, \cdots, \underbrace{1,1, \cdots, 1}_{n-1 \uparrow}, n, \cdots$ has the sum of its first 2007 terms as

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

9.3898 .

In the sequence, from number 1 to $n$ there are $n+\frac{n(n-1)}{2}$ terms, thus, from number 1 to 62 there are $62+\frac{62 \times(62-1)}{2}$ $=1953$ terms, followed by 54 ones.
Therefore, the sum of the first 2007 terms is
$\begin{array}{l} (1+2+\cdots+62)+(1+2+\cdots+61)+54 \\ =3898 \end{array}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.