Równanie rekurencyjne

[Albatross201]

Rozwiazac rownanie rekurencyjne: \(\displaystyle{ a_{0}=1,a_{1}=8,a_{n}=4a_{n-1}+4a_{n-2}, n \ge 2}\)
\(\displaystyle{ a_{n+2}=4a_{n+1}+4a_{n}}\)
\(\displaystyle{ r^{2}-4r+4=0}\)
\(\displaystyle{ r_{1}=2-2 \sqrt{2},r_{2}=2+2 \sqrt{2}}\)

\(\displaystyle{ a_{0}=c_{1}r_{1}^{0}+c_{2}r_{2}^{0}}\)
\(\displaystyle{ a_{1}=c_{1}r_{1}^{1}+c_{2}r_{2}^{1}}\)

\(\displaystyle{ 1=c_{1}+c_{2}}\)
\(\displaystyle{ 8=(2-2 \sqrt{2} )c_{1}+(2+2 \sqrt{2} )c_{2}}\)

\(\displaystyle{ c_{1}= \frac{2-3 \sqrt{2} }{4}}\)
\(\displaystyle{ c_{2}= \frac{2+3 \sqrt{2} }{4}}\)
Moja odp:
\(\displaystyle{ \frac{2-3 \sqrt{2} }{4}(2-2 \sqrt{2} )^{n}+\frac{2+3 \sqrt{2} }{4}(2+2 \sqrt{2} )^{n}}\)
Odp w ksiazce:
\(\displaystyle{ -\frac{1}{4}(2-2 \sqrt{2} )^{n+1}+ \frac{6+2 \sqrt{2} }{4}(2+2 \sqrt{2} )^{n}}\)

Gdzie mam blad?

[tometomek91]

\(\displaystyle{ a_{n+2}=4a_{n+1}+4a_{n}}\)
\(\displaystyle{ r^{2}-4r+4=0}\)

-4.

-- 7 kwi 2010, o 20:00 --

...i tak nie wpłyneło na dalsze rozwiązywanie

poza tym nie widzę błędu.