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Get the recurrence formula of $$U_n=2(-3)^n-5n(-3)^n$$ For $$n \geq 1$$

What am I supposed to do with this condition $n\geq 1$?

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It means that you can assume $n\geq1$ in all your calculations, i.e. you don't need to calculate $U_0$, $U_{-1}$ etc. –  Chris Taylor Jul 5 '12 at 10:01
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3 Answers

up vote 1 down vote accepted

$U_{n}=(-3)^{n}(2-5(n))$

$U_{n+1}=(-3)^{n+1}(2-5(n+1)) = 3(-3)^n(2-5n-5) $

$U_{n+1}= -3U_{n}+5(-3)^{n+1}$

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$U_{n}=(-3)^{n}(2-5(n))=(-3)^{(n-1)+1}(2-5((n-1)+1))=-3(-3)^{(n-1)}(2-5(n-1)-5)=-3.(-3)^{(n-1)}(2-5(n-1))+15(-3)^{(n-1)}=-3U_{n-1}+15(-3)^{n-1}=-3U_{n-1}+5(-3)^{n}$.

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"The" recurrence formula is not the best wording, since there are many recurrence formulas satisfied by the sequence $(U_n)$.

By the technique of avatar and lab bhattacharjee, we have $$U_{n}=-3U_{n-1}+5(-3)^{n},\tag{$1$} $$ and therefore $$U_{n+1}=-3U_n+5(-3)^{n+1}.\tag{$2$}$$ Multiply $(1)$ by $-3$, and subtract from $(2)$. We obtain the linear recurrence with constant coefficients $$U_{n+1}+6U_n+9U_{n-1}=0.\tag{$3$}$$ To get our exact sequence, we need to write down two initial conditions. For example we can use $U_1=9$, $U_2=-72$.

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