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2006 AIME II Problems/Problem 11

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Problem

A sequence is defined as follows a_1=a_2=a_3=1, and, for all positive integers n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. Given that a_{28}=6090307, a_{29}=11201821, and a_{30}=20603361, find the remainder when \sum^{28}_{k=1} a_k is divided by 1000.

Solution

Define the sum as s. Since a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}, the sum will be:

\begin{align*}s &= a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\&= a_{28} + \left(\sum^{30}_{k=4} a_{k} - \su...

Thus s = \frac{a_{28} + a_{30}}{2}, and a_{28},\,a_{30} are both given; the last four digits of their sum is 3668, and half of that is 1834. Therefore, the answer is \boxed{834}.

See also

2006 AIME II (Problems • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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