Câu hỏi 38

Ta có \({u_{n + 1}} = 4{u_n} - 3{u_{n - 1}}\).

+) \({u_2} = 4{u_1} - 3{u_0} = 2022 = {u_0} + {3^0} + {3^1}\)

Tương tự \({u_3} = 4{u_2} - 3{u_1} = 2031 = {u_0} + {3^0} + {3^1} + {3^2}\)

               \({u_4} = 4{u_3} - {u_2} = 2058 = {u_0} + {3^0} + {3^1} + {3^2} + {3^3}\)  

Suy ra \({u_n} = {u_0} + {3^0} + {3^1} + {3^2} + ... + {3^{n - 1}}\).

Ta có \({3^0} + {3^1} + ... + {3^{n - 1}} = 1.\dfrac{{1 - {3^n}}}{{1 - 3}} = \dfrac{{{3^n} - 1}}{2}\).

\( \Rightarrow {u_n} = 2018 + \dfrac{{{3^n} - 1}}{2} = \dfrac{{4035}}{2} + \dfrac{1}{2}{3^n}\).

Vậy \(\lim \dfrac{{{u_n}}}{{{3^n}}} = \lim \dfrac{{\dfrac{{4035}}{2} + \dfrac{1}{2}{3^n}}}{{{3^n}}} = \lim \left( {\dfrac{{4035}}{{{{2.3}^n}}} + \dfrac{1}{2}} \right) = \dfrac{1}{2}.\)

Chọn C.