Recent content by dragonkiller1

Oh got it. Thanks :)

I've showed the base step for n=1 For the inductive step: Suppose P(k) is true for some k \begin{bmatrix} x_{k}\\ x_{k-1} \end{bmatrix} = \begin{bmatrix} 5 &-6 \\ 1 & 0 \end{bmatrix}^{k-1}\begin{bmatrix} 1\\0 \end{bmatrix} Then for n=k+1 \begin{bmatrix} x_{k+1}\\ x_{k}...

x_{xn-1}= 5_{xn-1} - 6_{xn-2};for \ n≥2 \\ x_{1}=1 \\ x_{0}=0 prove by induction that: \begin{bmatrix} x_{n}\\ x_{n-1} \end{bmatrix} = \begin{bmatrix} 5 &-6 \\ 1 & 0 \end{bmatrix}^{n-1}\begin{bmatrix} 1\\0 \end{bmatrix}