recurrence relation – math.stackexchange.com

recurrence relation – math.stackexchange.com

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I didn't do a lot of maths in my career, and we asked me to solve the following recurrence relation : $$a_{n} = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$$ with $a_0 = 2$, $a_1 = 3$ and $a_2 = 1$ ...

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