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How do you find a formula for the nth term of the sequence that satisfies: $x_n$$_+$$_1$=$-4x_n$-$4x_n$$_-$$_1$?

$x_0 = 1$, $x_1 = 0$

I substituted $x_n$ = $x^n$ then found that the roots of the quadratic are coincident roots, -2 repeated but I don't know what to do know.

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1 Answer 1

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The characteristic equation is $x^2+4x+4=0$, so it has a double root in $-2$.

Therefore, we have $x_n=a(-2)^n+bn(-2)^n$.

For $n=0$, we have $a=1$, for $n=1$, we have $0=-2a-2b$, so $b=-1$.

We have $x_n=(1-n)(-2)^n$.

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  • $\begingroup$ Thank you, I missed out the "n" for b"n" that's why I didn't get it. $\endgroup$ Dec 3, 2015 at 14:51

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