2008 iTest Problems/Problem 69

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Problem

In the sequence in the previous problem, how many of $u_1,u_2,u_3,\ldots, u_{2008}$ are pentagonal numbers?

Note: The sequence is $1,2,5,6,9,12 \cdots,$ where the first term is 1, the next 2 terms are congruent to 2 modulo 3, the next 3 terms are congruent to 3 modulo 3, and so on.

Solution

The first few pentagonal numbers are $1,5,12,22 \cdots,$ which seems to match the last term of each grouping. Notice that the common differences are the same.


By checking with the formula $p_n = \tfrac{3n^2 - n}{2}$ (or using a visual representation of pentagonal numbers), we confirm that the first difference is always one more than a multiple of 3, and the second difference is 3. This confirms that the last number of each grouping are pentagonal numbers. In addition, the last number of each grouping is term number $\tfrac{n(n+1)}{2}.$


With this in mind, the highest $n$ such that $\tfrac{n(n+1)}{2} < 2008$ is $62$, so there are $\boxed{62}$ pentagonal numbers in the sequence.

See Also

2008 iTest (Problems)
Preceded by:
Problem 68
Followed by:
Problem 70
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