Difference between revisions of "2006 AIME A Problems/Problem 11"

 
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== Problem ==
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#REDIRECT [[2006 AIME I Problems/Problem 11]]
A sequence is defined as follows <math> a_1=a_2=a_3=1, </math> and, for all positive integers <math> n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. </math> Given that <math> a_{28}=6090307, a_{29}=11201821, </math> and <math> a_{30}=20603361, </math> find the remainder when <math> \displaystyle \sum^{28}_{k=1} a_k </math> is divided by 1000.
 
 
 
== Solution ==
 
 
 
== See also ==
 
*[[2006 AIME II Problems]]
 

Latest revision as of 14:15, 23 August 2009

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