Bài 6 (1.0 điểm). Cho dãy số $\left(u_n\right)$ biết $\left\{\begin{array}{l}u_1=5, u_2=13 \\ u_{n+2}=8 u_{n+1}-15 u_n+3 \cdot 4^n, \forall n \in \mathbb{N}^*\end{array}\right.$. Tính $\lim \frac{u_n}{5^{n+1}}$.

Xét dãy số `v_n = u_n - 3.4^{n-1}`

`v_1 = u_1 - 3 = 2`

`v_2 = u_2 - 12 = 1`

`u_{n+2} = 8u_{n+1} - 15u_n + 3.4^n`

`u_{n+2} - 3.4^{n+1} = 8(u_{n+1} - 3.4^n) - 15(u_n - 3.4^{n-1})`

`v_{n+2} = 8v_{n+1} - 15v_n`

Phương trình đặc trưng: `x^2 - 8x + 15 = 0`

`x_1 = 3, x_2 = 5`

`v_n = A.3^n + B.5^n`

`v_1 = 3A + 5B = 2`

`v_2 = 9A + 25B = 1`

`A = -\frac{1}{6}, B = \frac{1}{2}`

`v_n = \frac{5^n - 3^n}{2}`

`u_n = \frac{5^n - 3^n}{2} + 3.4^{n-1}`

`\lim \frac{u_n}{5^n+1} = \lim \frac{5^n - 3^n + 3.4^{n-1}}{2(5^n+1)} = \lim \frac{1 - (\frac{3}{5})^n + 3(\frac{4}{5})^n}{2(1+\frac{1}{5^n})} = \frac{1}{2}`