2006 AIME II Problems/Problem 11
Define the sum as . Since , the sum will be:
Thus , and are both given; the last four digits of their sum is , and half of that is . Therefore, the answer is .
Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:
Adding all the residues shows the sum is congruent to mod 1000.
Solution 3 (some guessing involved)
All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given and , so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some such that . From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that , at least for the first few terms. From this, we have that .
Solution by zeroman
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