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 $n\geq 1$?
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 $n\geq 1$?
$U_{n}=(-3)^{n}(2-5(n))$
$U_{n+1}=(-3)^{n+1}(2-5(n+1)) = 3(-3)^n(2-5n-5) $
$U_{n+1}= -3U_{n}+5(-3)^{n+1}$
$U_{n}=(-3)^{n}(2-5(n))=(-3)^{(n-1)+1}(2-5((n-1)+1))=-3(-3)^{(n-1)}(2-5(n-1)-5)=-3.(-3)^{(n-1)}(2-5(n-1))+15(-3)^{(n-1)}=-3U_{n-1}+15(-3)^{n-1}=-3U_{n-1}+5(-3)^{n}$.
"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 $U_1=9$, $U_2=-72$.