Lập dãy $\left ( \alpha _{\,\text{n}} \right )$ xác định bởi: $\alpha _{\,0}= 0,\,\alpha _{\,\text{n}+ 1}= \alpha _{\,\text{n}}+ 1,\,\text{n}= 0,\,1,\,2,\,...$ . Chứng minh rằng với $\text{P}\left ( x^{\,2}+ 1 \right )= \left [ \text{P}\left ( x \right )- 6 \right ]^{\,2}+ 7\,\,,\,\,\forall\,x \in \mathbb{R}$ thì:
$$\text{P}\left ( \alpha _{\,\text{n}} \right )= \alpha _{\,\text{n}}+ 6$$
