Giải và nêu bước làm

Ta có: `u_(n+1)=4u_n+u_(n-1) \ forall \ n in NN^**`

`<=>` `u_(n+1)+(-2+sqrt5)u_n=(2+sqrt5)u_n+u_(n-1)`

`<=>` `u_(n+1)+(-2+sqrt5)u_n=(2+sqrt5)[u_n+(-2+sqrt5)u_(n-1)]`

Đặt `v_n=u_n+(-2+sqrt5)u_(n-1)`

`=>` `(v_n): \ {(v_1=sqrt5),(v_(n+1)=(2+sqrt5)v_n \ forall \ n in NN^**):}`

`=>` `v_n=sqrt5*(2+sqrt5)^(n-1) \ forall \ n in NN^**`

`=>` `u_n+(-2+sqrt5)u_(n-1)=sqrt5*(2+sqrt5)^(n-1) \ forall \ n in NN^**`

`<=>` `u_n=(2-sqrt5)u_(n-1)+sqrt5*(2+sqrt5)^(n-1) \ forall \ n in NN^**`

`<=>` `u_n-1/2*(2+sqrt5)^n=(2-sqrt5)*[u_(n-1)-1/2*(2+sqrt5)^(n-1)]`

Đặt `w_n=u_n-1/2*(2+sqrt5)^n`

`=>` `{(w_0=1/2),(w_1=(2-sqrt5)/2),(w_n=(2-sqrt5)w_(n-1) \ forall \ n in NN; n >= 2):}`

`=>` `w_n=1/2*(2-sqrt5)^n \ forall \ n in NN`

`=>` `u_n=1/2*(2-sqrt5)^n+1/2*(2+sqrt5)^n \ forall \ n in NN`