Exame de´Algebra Linear e Geometria Anal´ıtica

Portanto:
( ) ( )
˜x1 5˜x0
= ,
ỹ 1 10ỹ 0
( ) ( ) ( )
˜x2 5˜x1 52˜x
= = 0
ỹ 2 10ỹ 1 10 2 , · · · ỹ 0
( ) ( )
˜xn 5n˜x
= 0
ỹ n 10 n ỹ 0
Mas ˜x 0 = 2x √5 0+y 0
= √ 3 5
, ỹ 0 = x 0−2y √5 0
= √ −1
5
. Portanto:
{
˜xn = 2x n+y √5 n
= 5 n √ 3 5
ỹ n = xn−2yn √
5
= 10 n √ −1
5
e resolvendo em ordem a x n e y n obtemos:
x n = 2 × 5 n−1 (3 − 2 n−1 ), y n = 5 n−1 (3 + 4 × 2 n−1 )
3