Cho dãy số (un) có: u1 = 4; u(n+1) = 3un + 4; với mọi n = 1, 2, 3
Câu hỏi:
Cho dãy số (un) có: \(\left\{ \begin{array}{l}{u_1} = 4\\{u_{n + 1}} = 3{u_n} + 4;\forall n = 1,2,3,...\end{array} \right.\).
Tính \(\lim \frac{{{u_n} + {2^n}}}{{{2^{n + 1}} - 3}}\).
Trả lời:
\(\left\{ \begin{array}{l}{u_1} = 4\\{u_{n + 1}} = 3{u_n} + 4;\forall n = 1,2,3,...\end{array} \right.\)
⇔ \(\left\{ \begin{array}{l}{u_1} = 4\\{u_{n + 1}} + 2 = 3{u_n} + 6\end{array} \right.\)
⇔ \(\left\{ \begin{array}{l}{u_1} = 4\\{u_{n + 1}} + 2 = 3\left( {{u_n} + 2} \right)\end{array} \right.\)
Đặt vn = un + 2 nên suy ra: v1 = u1 + 2 = 4 + 2 = 6
Từ đó ta có:
\(\left\{ \begin{array}{l}{v_1} = 6\\{v_{n + 1}} = 3{v_n}\end{array} \right.\)
Suy ra: vn = 6.3n–1
⇒ un = 6.3n–1 – 2
Ta có: \(\lim \frac{{{u_n} + {2^n}}}{{{2^{n + 1}} - 3}} = \lim \frac{{{{6.3}^{n - 1}} - 2 + {2^n}}}{{{2^{n + 1}} - 3}} = \lim \frac{{{{2.3}^n} - 2 + {2^n}}}{{{{2.2}^n} - 3}}\)
\( = \lim \frac{{2 - \frac{2}{{{3^n}}} + {{\left( {\frac{2}{3}} \right)}^n}}}{{2.{{\left( {\frac{2}{3}} \right)}^n} - \frac{3}{{{3^n}}}}} = \frac{2}{0} = + \infty \).
