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My task: $a_n=6a_{n-1}-9a_{n-2}-8$ for $n\geq2$, $a_0=0$, $a_1=1$

My solution $x^{2}-6x+9$

$\Delta=0$

$x_0=3 $

So I am gonna use following formula: $a_n=ar^{n}+bnr^{n}$

$a_n=a*(3)^{n}+bn*3^{n}$

$-8$ is the problem, so I am looking for $c$ that $b_n:=a_n+c\implies b_n=6b_{n-1}-9b_{n-2}$

$$b_n=6(b_{n-1}-c)-9(b_{n-2}-c)-8+c=6b_{n-1}-9b_{n-2}-8+6c$$ I am setting $c=\frac{4}{3}$ so

$$b_n=6b_{n-1}-9b_{n-2}\implies\exists a,\,b:\,b_n=a*3^{n}+bn*3^{n}.$$From $b_0=\frac{4}{3},\,b_1=\frac{7}{3}$, after finding $a,\,b$. Then $a_n=b_n-\frac{1}{2}$.

$$a=\frac{4}{3}$$ $$b=-\frac{5}{9}$$ $b_2=22$

$a_2=22-\frac{4}{3}=\frac{62}{3}$

Actual $a_2=-2$

So $a_2$ from $b_n$ method is not equal to actual $a_n$.

Can I use this $b_n$ method if delta equals 0? or should $c=-\frac{4}{3}$?

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  • $\begingroup$ How did you compute $b_0$ and $b_1$? $\endgroup$ – Exodd Jan 27 at 21:03
  • $\begingroup$ $$b_0=a_0+\frac{4}{3}=\frac{4}{3}$$ $$b_1=a_1+\frac{4}{3}=\frac{7}{3}$$ Forgot to edit $\endgroup$ – Arnolt Infern Kitler Jan 27 at 21:09
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I think that your choice of $c$ is wrong From $$b_n=6(b_{n-1}-c)-9(b_{n-2}-c)-8+c=6b_{n-1}-9b_{n-2}-8+4c$$

this gives you $c=2$ and $b_n=a_n+2$

$$b_0=2, b_1=3$$ after solving $$b_n=a*3^{n}+bn*3^{n}$$ you get $a=2 ,b=-1$

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