Cho dãy số (un) có: u1 = 4; u(n+1) = 3un + 4; với mọi n = 1, 2, 3

Cho dãy số (u_n) có: $\left\{ \begin{array}{l}{u_1} = 4\\{u_{n + 1}} = 3{u_n} + 4;\forall n = 1,2,3,...\end{array} \right.$.

Tính $\lim \frac{{{u_n} + {2^n}}}{{{2^{n + 1}} - 3}}$.

$\left\{ \begin{array}{l}{u_1} = 4\\{u_{n + 1}} = 3{u_n} + 4;\forall n = 1,2,3,...\end{array} \right.$

⇔ $\left\{ \begin{array}{l}{u_1} = 4\\{u_{n + 1}} + 2 = 3{u_n} + 6\end{array} \right.$

⇔ $\left\{ \begin{array}{l}{u_1} = 4\\{u_{n + 1}} + 2 = 3\left( {{u_n} + 2} \right)\end{array} \right.$

Đặt v_n = u_n + 2 nên suy ra: v₁ = u_1­ + 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: v_n = 6.3^n–1

⇒ u_n = 6.3^n–1 – 2

Ta có: $\lim \frac{{{u_n} + {2^n}}}{{{2^{n + 1}} - 3}} = \lim \frac{{{{6.3}^{n - 1}} - 2 + {2^n}}}{{{2^{n + 1}} - 3}} = \lim \frac{{{{2.3}^n} - 2 + {2^n}}}{{{{2.2}^n} - 3}}$

$= \lim \frac{{2 - \frac{2}{{{3^n}}} + {{\left( {\frac{2}{3}} \right)}^n}}}{{2.{{\left( {\frac{2}{3}} \right)}^n} - \frac{3}{{{3^n}}}}} = \frac{2}{0} = + \infty$.