Cho dãy số $(u_n)$ xác định bởi:\begin{cases}u_1= 5\\ u_{n+1}=2u_n+3-3n, \forall n \in N \end{cases}Xác định công thức tổng quát của dãy. Xét xem dãy tăng hay giảm?
Ta chứng minh mệnh đề trên bằng quy nạp:
$u_1=5=2^1+3.1,         u_2=2u_1+3-6=2.5=10$
Dự đoán  $u_n=2^n+3n        (1)$
+ Với $n=1: u_1=2^1+3.1=5,  (1)$ đúng.   
Giả sử $(1)$ đúng với $n=k$.
+ $n=k:  u_k=2^k+3k$
   $u_{k+1}=2u_k+3-3k=2(2^k+3k)+3-3k=2^{k+1}+3(k+1)$
Vậy $(1)$ đúng với $n=k+1$. Do đó, theo nguyên lí quy nạp $(1)$ đúng $\forall n \in N$.
ĐS:  $u_n=2^n+3n$

b) Xét hiệu $H=u_{n+1}-u_n=2^{n+1}+3(n+1)-2^n-3n=2^n+3>0  \forall n\in N$.
$H>0$ chứng tỏ $(u_n)$ là dãy số tăng.

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