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The recurrence relation  g(n)=3g(n−1)+2[g(n−2)+g(n−3)+g(n−4)+⋯+g(2)+g(1)] can be simplified to  g(n)=αg(n−1)+βg(n−2) . The value of  α+β is

 Jul 4, 2023
 #1
avatar+33616 
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Like so:

 

 Jul 4, 2023
 #2
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That is a very nice solution, alan!
 Jul 4, 2023

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